property assignment expected.ts(1136) angular

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Property assignment expected. Unable to build project angular 5 #12147

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Home » Introduction and Basics » Typescript object destructuring results in property assignment expected

Typescript object destructuring results in property assignment expected

Introduction.

Typescript is a popular programming language that adds static typing to JavaScript. It provides developers with the ability to catch errors during development and improve code quality. However, like any programming language, Typescript has its own set of challenges and issues that developers may encounter. One common issue is the “Typescript object destructuring results in property assignment expected” error. In this article, we will explore this error and provide solutions with examples.

The “Typescript object destructuring results in property assignment expected” error occurs when you try to destructure an object but forget to provide a default value for the property being destructured. Let’s take a look at an example:

In this example, we are trying to destructure the “person” object and assign the value of the “name” property to a variable called “name”. However, since we did not provide a default value for the “name” property, Typescript throws the “Typescript object destructuring results in property assignment expected” error.

To solve this error, we need to provide a default value for the property being destructured. We can do this by using the assignment operator (=) and specifying a default value after the property name. Let’s modify our previous example to include a default value:

In this modified example, we have provided a default value of an empty string (“”) for the “name” property. Now, even if the “name” property is not present in the “person” object, the destructuring assignment will not throw an error.

The “Typescript object destructuring results in property assignment expected” error can be easily solved by providing a default value for the property being destructured. By doing so, we ensure that the destructuring assignment does not throw an error even if the property is not present in the object. This helps in writing more robust and error-free Typescript code.

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• Property or signature expected error in TypeScript

Last updated: Feb 28, 2024 Reading time · 2 min

# Property or signature expected error in TypeScript

The "Property or signature expected" error occurs when we have a syntax error when declaring a type or interface.

To solve the error, make sure to separate the properties and types by a colon and wrap properties that contain hyphens or spaces in quotes.

Here are 3 examples of how the error occurs.

The cause of the error in the first example is that the property name in the interface contains a hyphen.

# Wrap hyphenated properties in quotes

If a property name in an interface, type or object contains a hyphen, we have to wrap it in quotes.

Note that we had to wrap the property in quotes both in the interface and when declaring an object of type Employee .

The same is the case with spaces (and most other separators).

# Wrap properties that contain spaces in quotes

If an object, type or interface property contains a space, wrap it in quotes.

# Separate properties and types by a colon

Another common cause of the error is mistakenly separating the property and the type by an equal sign or any other symbol that isn't a colon.

Always make sure to separate the property name and the type by a colon.

Note that you can't provide default values in types and interfaces , because TypeScript only helps us catch bugs during development.

All of the types and interfaces you have defined will be removed when your code is compiled to JavaScript.

# Additional Resources

You can learn more about the related topics by checking out the following tutorials:

• How to export Interfaces and Types in TypeScript
• An interface can only extend an object type or intersection of object types
• Property is private and only accessible within class in TS
• Property is missing in type 'X' but required in type 'Y'
• Property has no initializer and is not definitely assigned in the constructor
• Property is incompatible with index signature in TypeScript

Borislav Hadzhiev

Web Developer

Copyright © 2024 Borislav Hadzhiev

Angular Property Assignment Error

Breakdown of the Error:

• Property: This refers to a specific variable or member within a class or component.
• Initializer: An initializer is a value assigned to a property when it's declared. For example, let x = 5; initializes x with the value 5.
• Definitely assigned: In TypeScript, a property is considered definitely assigned if its value is guaranteed to be set before it's used. This ensures that the property won't be accessed when it's still undefined.

The Problem:

When you encounter this error, it means that you've declared a property in your component, but you haven't provided an initial value for it, and the compiler cannot guarantee that it will be assigned a value before it's used. This can lead to runtime errors if you try to access the property while it's still undefined.

Common Causes:

• Missing initializer: You simply forgot to provide an initial value when declaring the property.
• Conditional assignment: The property might be assigned a value only under certain conditions, and the compiler cannot determine if those conditions will always be met.
• Asynchronous operations: If the property's value is set asynchronously (e.g., using a service or HTTP request), the compiler might not be able to guarantee that it's assigned before it's used.

To address this error, you need to ensure that the property is definitely assigned before it's used. Here are some common solutions:

• class MyComponent { name : string = "John Doe" ; }

Assign the property in the constructor: If you need to set the property's value based on other factors or during initialization, you can assign it in the constructor:

Use a conditional check: If the property's value depends on a condition, you can use a conditional check to ensure that it's assigned before accessing it:

• class MyComponent { name?: string ; }

Understanding "Property '...' has no initializer and is not definitely assigned in the constructor" in Angular

Error Breakdown:

• Property: A variable within a class or component.
• Initializer: A value assigned to a property when it's declared.
• Definitely assigned: The property's value is guaranteed to be set before it's used.

Explanation: The error occurs because name is declared but doesn't have an initial value. TypeScript requires properties to be definitely assigned to prevent potential runtime errors.

Initializer: Provide an initial value when declaring the property:

Constructor Assignment: Assign the property in the constructor:

Conditional Check: Assign the property based on conditions:

Understanding Angular Property Assignment Error

This error typically arises when you try to assign a value to a property that doesn't exist or is readonly.

Explanation: The name property is declared as readonly , meaning its value cannot be changed after initialization. Trying to assign a new value to it causes the error.

Remove readonly : If you need to modify the property, remove the readonly modifier:

Create a Setter: Define a setter method to modify the property indirectly:

• If you're unsure whether a property will be defined, use optional chaining to safely access it: class MyComponent { user : User | null ; constructor ( ) { // ... } getName ( ) { return this .user?.name || 'Unknown' ; } }

Nullish Coalescing Operator (ES2020):

• Provide a default value for a property if it's null or undefined: class MyComponent { name : string | null ; constructor ( ) { // ... } getDisplayName ( ) { return this .name ?? 'Default Name' ; } }

Default Parameters:

• Set a default value for a property in the constructor: class MyComponent { constructor ( public name: string = 'Default Name' ) {} }

Dependency Injection:

• Inject a service that provides the necessary data: class MyComponent { constructor ( private userService: UserService ) {} ngOnInit ( ) { this .userService.getUser().subscribe( user => { this .user = user; }); } }

Alternative Methods for "Angular Property Assignment Error"

Conditional Assignment:

• Check if the property is defined before assigning a value: class MyComponent { name : string | undefined ; updateName ( newName: string ) { if ( this .name !== undefined ) { this .name = newName; } } }

Template-Driven Forms:

• Use two-way data binding to automatically update component properties: < input [( ngModel )]= "name" >

Reactive Forms:

• Create a FormGroup and use form controls to manage property values: import { FormControl } from '@angular/forms' ; class MyComponent { nameControl = new FormControl( '' ); }

Lifecycle Hooks:

• Use ngOnInit to initialize properties based on asynchronous data: class MyComponent { name : string ; ngOnInit ( ) { this .userService.getUser().subscribe( user => { this .name = user.name; }); } }

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How to dynamically assign properties to an object in TypeScript

Editor’s note: This article was updated on 6 October 2023, introducing solutions like type assertions and the Partial utility type to address the TypeScript error highlighted.

JavaScript is a dynamically typed language, meaning that a variable’s type is determined at runtime and by what it holds at the time of execution. This makes it flexible but also unreliable and prone to errors because a variable’s value might be unexpected.

TypeScript, on the other hand, is a statically typed version of JavaScript — unlike JavaScript, where a variable can change types randomly, TypeScript defines the type of a variable at its declaration or initialization.

Dynamic property assignment is the ability to add properties to an object only when they are needed. This can occur when an object has certain properties set in different parts of our code that are often conditional.

In this article, we will explore some ways to enjoy the dynamic benefits of JavaScript alongside the security of TypeScript’s typing in dynamic property assignment.

Consider the following example of TypeScript code:

This seemingly harmless piece of code throws a TypeScript error when dynamically assigning name to the organization object:

See this example in the TypeScript Playground .

The source of confusion, perhaps rightly justified if you’re a TypeScript beginner, is: how could something that seems so simple be such a problem in TypeScript?

The TL;DR of it all is that if you can’t define the variable type at declaration time, you can use the Record utility type or an object index signature to solve this. But in this article, we’ll go through the problem itself and work toward a solution that should work in most cases.

The problem with dynamically assigning properties to objects

Generally speaking, TypeScript determines the type of a variable when it is declared. This determined type stays the same throughout your application. There are exceptions to this rule, such as when considering type narrowing or working with the any type, but otherwise, this is a general rule to remember.

In the earlier example, the organization object is declared as follows:

There is no explicit type assigned to this variable, so TypeScript infers a type of organization based on the declaration to be {} , i.e., the literal empty object.

If you add a type alias, you can explore the type of organization :

See this in the TypeScript Playground .

When you then try to reference the name prop on this empty object literal:

You receive the following error:

There are many ways to solve the TypeScript error here. Let’s consider the following:

Solution 1: Explicitly type the object at declaration time

This is the easiest solution to reason through. At the time you declare the object, go ahead and type it, and assign all the relevant values:

This eliminates any surprises. You’re clearly stating what this object type is and rightly declaring all relevant properties when you create the object.

However, this is not always feasible if the object properties must be added dynamically, which is why we’re here.

Solution 2: Use an object index signature

Occasionally, the properties of the object truly need to be added at a time after they’ve been declared. In this case, you can use the object index signature, as follows:

When the organization variable is declared, you can explicitly type it to the following: {[key: string] : string} .

You might be used to object types having fixed property types:

However, you can also substitute name for a “variable type.” For example, if you want to define any string property on obj :

Note that the syntax is similar to how you’d use a variable object property in standard JavaScript:

The TypeScript equivalent is called an object index signature.

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Moreover, note that you could type key with other primitives:

Solution 3: Use the Record Utility Type

The Record utility type allows you to constrict an object type whose properties are Keys and property values are Type . It has the following signature: Record<Keys, Type> .

In our example, Keys represents string and Type . The solution here is shown below:

Instead of using a type alias, you can also inline the type:

Solution 4: Use the Map data type

A Map object is a fundamentally different data structure from an object , but for completeness, you could eliminate this problem if you were using Map .

Consider the starting example rewritten to use a Map object:

With Map objects, you’ll have no errors when dynamically assigning properties to the object:

This seems like a great solution at first, but the caveat is your Map object is weakly typed. You can access a nonexisting property and get no warnings at all:

See the TypeScript Playground .

This is unlike the standard object. By default, the initialized Map has the key and value types as any — i.e., new () => Map<any, any> . Consequently, the return type of the s variable will be any :

When using Map , at the very least, I strongly suggest passing some type information upon creation. For example:

s will still be undefined, but you won’t be surprised by its code usage. You’ll now receive the appropriate type for it:

If you truly don’t know what the keys of the Map will be, you can go ahead and represent this at the type level:

And if you’re not sure what the keys or values are, be safe by representing this at the type level:

Solution 5: Consider an optional object property

This solution won’t always be possible, but if you know the name of the property to be dynamically assigned, you can optionally provide this when initializing the object as shown below:

If you don’t like the idea of using optional properties, you can be more explicit with your typing as shown below:

Solution 6: Leveraging type assertions

TypeScript type assertion is a mechanism that tells the compiler the variable’s type and overrides what it infers from the declaration or assignment. With this, we are telling the compiler to trust our understanding of the type because there will be no type verification.

We can perform a type assertion by either using the <> brackets or the as keyword. This is particularly helpful with the dynamic property assignment because it allows the properties we want for our object to be dynamically set because TypeScript won’t enforce them.

Let’s take a look at applying type assertions to our problem case:

Note that with type assertions, the compiler is trusting that we will enforce the type we have asserted. This means if we don’t, for example, set a value for organization.name , it will throw an error at runtime that we will have to handle ourselves.

Solution 7: Use the Partial utility type

TypeScript provides several utility types that can be used to manipulate types. Some of these utility types are Partial , Omit , Required , and Pick .

For dynamic property assignments, we will focus specifically on the Partial utility type. This takes a defined type and makes all its properties optional. Thus, we can initialize our object with any combination of its properties, from none to all, as each one is optional:

In our example with the Partial utility type, we defined our organization object as the type partial Org , which means we can choose not to set a phoneNumber property:

Grouping and comparing the options for adding properties in TypeScript

In this article, we explored the different options for setting properties dynamically in TypeScript. These options can be grouped together by their similarities.

Index/Key signatures

This group of options allows you to define the type of keys allowed without limiting what possible keys can exist. The options in this group include:

• Using an object index signature
• Using the Record utility type
• Using the Map data type (with key/value typing)

With these, we can define that our object will take string indexes and decide what types to support as values, like String , Number , Boolean , or Any :

See in TypeScript Playground .

Pro: The main benefit of these methods is the ability to dynamically add properties to an object while still setting expectations for the potential types of keys and values.

Con: The main disadvantage of this way of defining objects is that you can’t predict what keys our objects will have and so some references may or may not be defined. An additional disadvantage is that if we decide to define our key signature with type Any , then the object becomes even more unpredictable.

Conditional/Optional properties

This set of object assignment methods shares a common feature: the definition of optional properties. This means that the range of possible properties are known but some may or may not be set. The options in this group include:

• Using optional object properties
• Using the Partial utility type
• Using type assertions

See this example in the TypeScript Playground , or in the code block below:

Note: While these options mean that the possible keys are known and may not be set, TypeScript’s compiler won’t validate undefined states when using type assertions. This can lead to unhandled exceptions during runtime. For example, with optional properties and the Partial utility type, name has type string or undefined . Meanwhile, with type assertions, name has type string .

Pro: The advantage of this group of options is that all possible object keys and values are known.

Con: The disadvantage is that while the possible keys are known, we don’t know if those keys have been set and will have to handle the possibility that they are undefined.

Apart from primitives, the most common types you’ll have to deal with are likely object types. In cases where you need to build an object dynamically, take advantage of the Record utility type or use the object index signature to define the allowed properties on the object.

If you’d like to read more on this subject, feel free to check out my cheatsheet on the seven most-asked TypeScript questions on Stack Overflow, or tweet me any questions . Cheers!

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3 Replies to "How to dynamically assign properties to an object in TypeScript"

I know this is explicitly for TypeScript, and I think type declarations should always be first. But in general, you can also use a Map object. If it’s really meant to by dynamic, might as well utilize the power of Map.

Great suggestion (updated the article). It’s worth mentioning the weak typing you get by default i.e., with respect to Typescript.

Hi, thanks for your valuable article please consider ‘keyof type’ in TypeScript, and add this useful solution if you are happy have nice time

Leave a Reply Cancel reply

Strict Property Initialization in TypeScript

TypeScript 2.7 introduced a new compiler option for strict property initialization checks in classes. If the --strictPropertyInitialization flag is enabled, the type checker verifies that each instance property declared in a class either

• has a type that includes undefined ,
• has an explicit initializer , or
• is definitely assigned to in the constructor.

The --strictPropertyInitialization option is part of the family of compiler options that is enabled automatically when the --strict flag is set. As with all the other strict compiler options, you can set --strict to true and selectively opt out of strict property initialization checks by setting --strictPropertyInitialization to false .

Note that the --strictNullChecks flag must be set (either directly or indirectly via --strict ) in order for --strictPropertyInitialization to have any effect.

Alright, let's see strict property initialization checks in action. Without the --strictPropertyInitialization flag enabled, the following code type-checks just fine, but produces a TypeError at runtime:

The reason for the runtime error is that the username property holds the value undefined because there's no assignment to that property. Therefore, the call to the toLowerCase() method fails.

If we enable --strictPropertyInitialization , the type checker raises an error:

Let's look at four different ways we can properly type our User class to make the type error go away.

# Solution #1: Allowing undefined

One way to make the type error go away is to give the username property a type that includes undefined :

Now, it's perfectly valid for the username property to hold the value undefined . Whenever we want to use the username property as a string, though, we first have to make sure that it actually holds a string and not the value undefined , e.g. using typeof :

Alternatively, we can use optional chaining (the ?. operator) to only call the toLowerCase() method if the username property holds a non-nullish value. We can combine that with nullish coalescing (the ?? operator) to provide the fallback value:

# Solution #2: Explicit Property Initializer

Another way to make the type error go away is to add an explicit initializer to the username property. This way, the property holds a string value right away and is not observably undefined :

# Solution #3: Assignment in the Constructor

Perhaps the most useful solution is to add a username parameter to the constructor, which is then assigned to the username property. This way, whenever an instance of the User class is constructed, the caller has to provide the username as an argument:

We could simplify the User class by removing the explicit assignment to the class field and adding the public modifier to the username constructor parameter:

Note that strict property initialization requires each property to be definitely assigned in all possible code paths in the constructor. The following (contrived) example is therefore not type-correct because in some cases, we leave the username property uninitialized:

# Solution #4: Definite Assignment Assertion

If a class property neither has an explicit initializer nor a type including undefined , the type checker requires that property to be initialized directly within the constructor; otherwise, strict property initialization checks will fail. This is problematic if you want to initialize a property within a helper method or have a dependency injection framework initialize it for you. In these cases, you have to add a definite assignment assertion ( ! ) to that property's declaration:

By adding a definite assignment assertion to the username property, we're telling the type checker that it can expect the username property to be initialized, even if it cannot detect that on its own. It is now our responsibility to make sure the property is definitely assigned to after the constructor returns, so we have to careful; otherwise, the username property can be observably undefined and we're back to the TypeError at runtime.

This article and 44 others are part of the TypeScript Evolution series. Have a look!

Property has no initializer and is not definitely assigned

Option 1 - Initialize the property

Option 2 - make the property optional, doing this in an angular app.

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How to dynamically assign properties to an object in TypeScript

Introduction.

Consider the following example:

This seemingly harmless piece of code throws a TypeScript error on dynamically assigning name to the organization object.

An error is thrown when dynamically assigning a property to an object

The source of confusion, which is perhaps rightly justified if you’re a TypeScript beginner, is: how could something that seems so simple be such a problem in TypeScript?

The TL;DR of it all is that, if you cannot define the variable type at declaration time, you can use the Record utility type or an object index signature to solve this. But in this post, we’ll go through the problem itself and work toward a solution that should work in most cases.

Understanding the problem with dynamically assigning properties to objects

Generally speaking, TypeScript determines the type of a variable when it is declared, and this determined type doesn’t change, i.e., it stays the same all through your application.

There are exceptions to this rule, such as when considering type narrowing or working with any type, but this is a general rule to remember otherwise.

In the earlier example, the organization object is declared as follows:

Exploring the literal object type

When you then try to reference the name prop on this empty object literal:

TypeScript yells.

Property ‘name’ does not exist on type ‘ {}‘

When you understand the issue, the error does seem appropriate.

Let’s fix this.

Resolving the problem

There are numerous ways you can resolve the TypeScript error here. Let’s consider these:

Solution 1 : Explicitly type the object at declaration time

This is the easiest solution to reason through. At the time you declare the object, go ahead and type it. Furthermore, assign it all the relevant values.

This eliminates any surprises. You’re clearly stating what this object type is and rightly declaring all relevant properties when you create the object.

However, this is not always feasible if the object properties must be added dynamically , which is why we’re all here.

Solution 2 : Use an object index signature

Occasionally, the properties of the object truly need to be added at a time after they’ve been declared. In this case, you can use the object index signature, as follows:

At the time the organization variable is declared, you can go ahead and explicitly type it to the following {[key: string]: string}.

To explain the syntax further, you might be used to object types having fixed property types:

However, you can also substitute name for a “variable type”.

For example, if you want to define any string property on obj:

Note that the syntax is similar to how you’d use a variable object property in standard JavaScript:

The TypeScript equivalent is called an object index signature. Moreover, note that you could type key with other primitives:

Solution 3 : Use the Record utility type

The Record utility type allows you to constrict an object type whose properties are Keys and property values are Type. It has the following signature: Record<Keys, Type>.

In our example, Keys represent string and Type, string as well. The solution here is quite concise as shown below:

Instead of using a type alias, you can also inline the type:

Using the Record utility type

Apart from primitives, the most common types you’ll have to deal with are likely object types.

In cases where you need to build an object dynamically, take advantage of the Record utility type or use the object index signature to define the allowed properties on the object.

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Property assignment expected. ts(1136)

I am following book Learning React, chapter 13 page 158, to first create-react-app helloworld and create HelloWorld.js

error message: at <div , I got “Property assignment expected.ts(1136)” at className, I got 2 errors Unreachable code detected.ts(7027) ‘;’ expected.ts(1005)

package.json { “name”: “helloworld”, “version”: “0.1.0”, “private”: true, “dependencies”: { “ @testing-library /jest-dom”: “^4.2.4”, “ @testing-library /react”: “^9.5.0”, “ @testing-library /user-event”: “^7.2.1”, “react”: “^16.14.0”, “react-dom”: “^16.14.0”, “react-scripts”: “3.4.3” }, “scripts”: { “start”: “react-scripts start”, “build”: “react-scripts build”, “test”: “react-scripts test”, “eject”: “react-scripts eject” }, “eslintConfig”: { “extends”: “react-app” }, “browserslist”: { “production”: [ “>0.2%”, “not dead”, “not op_mini all” ], “development”: [ “last 1 chrome version”, “last 1 firefox version”, “last 1 safari version” ] } }

I am using visual studio code Version: 1.50.0 Commit: 93c2f0fbf16c5a4b10e4d5f89737d9c2c25488a3 Date: 2020-10-07T06:01:33.073Z Electron: 9.2.1 Chrome: 83.0.4103.122 Node.js: 12.14.1 V8: 8.3.110.13-electron.0 OS: Linux x64 5.4.0-51-generic

Can you help? Thanks!

Hi. Your jsx (html) should be wrapped by parens ( () ) or nothing at all, not curly braces ( {} ). I think that’s what is causing the error.

Thanks a lot. That is it.

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'Property assignment expected' when using Object.entries

Hello, I'm having a hard time finding any answers on SO and hoping someone might be able to help me with this. I'm building a Next.JS app. My component is pulling in props as an array of objects. Inside useEffect I'm using Object.entries to iterate through key value pairs in my props object and running that value through a fetch function. What I'd like to do is append each response to the corresponding object inside my props array, but I'm getting TS error 'Property assignment expected' in VSCode, and 'Unexpected token' error in the browser. My code: useEffect(async () => { for (const [key, { address }] of Object.entries(props.artWorks)) { await fetch( \ https://api.mapbox.com/geocoding/v5/mapbox.places/${address}.json?limit=2&access_token=${process.env.MAPBOX_KEY}` ) .then((response) => response.json()) .then((result) => { return setLocaleState({ ...props.artWorks, `${key}`: result.features[0].center }) }) } }, [])`

The error is occurring at setState, \ ${key}` If I remove the template literal the function works, assigning a new array, 'key' to my props object and rewriting it on each iteration, as I expect it would. Obviously, this is not what I'm after, I'm trying to append each response to the corresponding object inside artWorks` I hope that's all clear. Been stuck on this for a couple days, any help would be greatly appreciate. Thank you!

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Fix – Property does not exist on type ‘{}’ in TypeScript

The “Property does not exist on type ‘{}’” error occurs in typescript when you try to modify, set or access an object property that does not exist. For example, if you have an object with only two properties and you try to access a third property that’s not contained in the object type. You can fix the above error by explicitly adding the object property or use a variable of the same type or key names. You can bypass or override TypeScript type checking whenever preferable, because is not strictly to be used but I do suggest that you don’t do this often when building a big project. Doing it too much will throw a typescript error “Property does not exist on type ‘object’”

From the above, suppose the ‘object1’ is a JSON data that is coming from a particular API (application programming interface) as a result of an HTTP request made when we run the application. Because you didn’t define the object yourself, what you can do with it is to read only the available properties. We have an empty object already defined, therefore we cannot assign values or access properties that does not exist on the object type.

The above first example illustrate how to define an object when you know all of it required property names and types of value that will be assign to it. Note that we were able to create a new empty object ‘obj’ of ‘Student’ because we marked all the object properties as optional with the question-mark symbol ‘?’ (it can be provided or remain null by default). That way, we can initialize the object as empty and later assign value to its properties, that’s the concept you need to understand.

We have included all the needed properties, therefore we will not get the “Property does not exist on type {} ” error again when we access any of it because they all exist in the object. So, the key concept to note once again is that computers don’t lie and whenever you encounter such type of error, it means some adjustment needs to be done in your code. 

If you don’t know the type of key, values, shape or structure of your object’s ahead of time, you have to use what’s called index signature as show below.

Note that the first object property above has been declared as '{[key: string]: any}' with the index signature syntax.  Remember to use this syntax when you don’t know how some of your object key and it value will look like. It’s also used when creating a regular variables and angular reactive-forms when it comes to defining separate input validators. The meaning of the index signature is that within the square bracket ‘[key:string]’ represent a key name. And the ‘:any’ represent the return value of the key declared so, ‘[key: string]: any’ means our object property will have a key of string and can be assigned a value or a function that returns any. If you want type safety, it will be good to specify each key type if you already know what value will be assigned to it.

Check also, how to fix error – JSX expressions must have one parent element in React

However, Typescript will throw “Error: Property 'id' of type 'number' is not assignable to 'string' index type 'string'.ts(2411)” if part of the object key is indexed with type and value of string and you declare another key of type number. You might think there is a bug with the type safety but I will recommend you learn something about union-type and index-signature first for clear understanding.

Summary: Property does not exist on type Object

Thanks for reading and if you have any suggestions or challenges applying above guide, let me know.

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Property assignment expected react jsx

I'm trying to check if my variable (item.label) contains the string 'ToDate'.

However, i am getting the error Property assignment expected on this line ({item.label} === 'ToDate') ? (

• Your arrow function brackets seem off, what you currently have would try and return an object containing a JSX element (With no property name) –  DBS Commented Aug 6, 2021 at 8:59

3 Answers 3

Simplifying what you currently have makes the issue clearer:

The outer () mean that this is trying to return an object {<ChipComponent />} without a property name, hence the property assignment error.

If you remove the inner {} it just will return either the JSX element or null:

You are also directly checking equality, rather than partial matching, you can use includes() to achieve that.

Complete example:

You can use indexOf as you are checking wether string has ToDate or not.

refer here to know more about.

You are not checking the equality in any case

whatever value this expression produces {item.label} gets appended and show in browser.

CODESANDBOX

You should write logic in the {} not outside of this.

If you are checking if a string contains another string then you can use includes .

Remember includes is case-sensitive search

const str = "LargeStringSearchFrom"; console.log(str.includes("Search"));

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