invalid left hand side in assignment expression useref

The left-hand side of assignment expression may not be an optional property access

Last updated: Feb 28, 2024 Reading time · 4 min

# The left-hand side of assignment expression may not be an optional property access

The error "The left-hand side of an assignment expression may not be an optional property access" occurs when we try to use optional chaining (?.) to assign a property to an object.

To solve the error, use an if statement that serves as a type guard instead.

Here is an example of how the error occurs.

We aren't allowed to use the optional chaining (?.) operator on the left-hand side of an assignment.

# Use an if statement as a type guard to solve the error

To solve the error, use an if statement as a type guard before the assignment.

We used the loose not equals operator (!=), to check if the variable is NOT equal to null and undefined .

This works because when compared loosely, null is equal to undefined .

The if block is only run if employee doesn't store an undefined or a null value.

This is similar to what the optional chaining (?.) operator does.

# Using the non-null assertion operator to solve the error

You might also see examples online that use the non-null assertion operator to solve the error.

The exclamation mark is the non-null assertion operator in TypeScript.

When you use this approach, you basically tell TypeScript that this value will never be null or undefined .

Here is an example of using this approach to set a property on an object.

In most cases, you should use a simple if statement that serves as a type guard as we did in the previous code sample.

# Avoiding the error with a type assertion

You can also use a type assertion to avoid getting the error. However, this isn't recommended.

The (employee as Employee) syntax is called a type assertion.

Type assertions are used when we have information about the type of a value that TypeScript can't know about.

We effectively tell TypeScript that the employee variable will have a type of Employee and not to worry about it.

This could go wrong if the variable is null or undefined as accessing a property on a null or an undefined value would cause a runtime error.

# Using the logical AND (&&) operator to get around the error

You can also use the logical AND (&&) operator to avoid getting the error.

The logical AND (&&) operator checks if the value to the left is truthy before evaluating the statement in the parentheses.

If the employee variable stores a falsy value (e.g. null or undefined ), the code to the right of the logical AND (&&) operator won't run at all.

The falsy values in JavaScript are: false , undefined , null , 0 , "" (empty string), NaN (not a number).

All other values are truthy.

However, this approach can only be used to assign a single property at a time if the value is not equal to null and undefined .

# The optional chaining operator should only be used when accessing properties

The optional chaining (?.) operator short-circuits if the reference is equal to null or undefined .

The optional chaining (?.) operator will simply return undefined in the example because employee has a value of undefined .

The purpose of the optional chaining (?.) operator is accessing deeply nested properties without erroring out if a value in the chain is equal to null or undefined .

However, the optional chaining operator cannot be used on the left-hand side of an assignment expression.

# Additional Resources

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

• How to Check the Type of a Variable in TypeScript
• Exclamation Mark (non-null assertion) operator in TypeScript
• The ?. operator (optional chaining) in TypeScript
• Declare and Type a nested Object in TypeScript
• How to Add a property to an Object in TypeScript
• Check if a Property exists in an Object in TypeScript
• The left-hand side of an arithmetic operation must be type 'any', 'number', 'bigint' or an enum type

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Jan 26, 2017 6:00:03 AM | JavaScript - ReferenceError: invalid assignment left-hand side

Today we examine the invalid assignment error, which is thrown, as the name implies, when code attempts to perform an invalid assignment somewhere.

Next on the list in our extensive JavaScript Error Handling series we're going to examine the Invalid Left-Hand Assignment error in greater detail. The Invalid Left-Hand Assignment error is a sub-object of ReferenceError and is thrown, as the name implies, when code attempts to perform an invalid assignment somewhere.

In this post we'll look at a few code examples to illustrate some common methods of producing an Invalid Left-Hand Assignment error, as well as examine how to handle this error when it rears its ugly head. Let the party begin!

The Technical Rundown

• All JavaScript error objects are descendants of the  Error  object, or an inherited object therein.
• The  ReferenceError  object is inherited from the  Error  object.
• The Invalid Left-Hand Assignment error is a specific type of ReferenceError object.

When Should You Use It?

As one of the simplest JavaScript errors to understand, the Invalid Left-Hand Assignment error appears in only a handful of situations in which code is attempting to pass an assignment incorrectly. While this is generally thought of as a syntactic issue, JavaScript defines this particular assignment error as a ReferenceError, since the engine effectively assumes an assignment to a non-referenced variable is being attempted.

The most common example of an Invalid Left-Hand Assignment error is when attempting to compare a value using a assignment operator (=), rather than using a proper comparison operator (== or ===). For example, here we're attempting to perform a basic comparison of the variable name with the values John or Fred. Unfortunately, we've made the mistake of using the assignment operator =, instead of a comparison operator such as == or ===:

try { var name = 'Bob'; if (name = 'John' || name = 'Fred') { console.log(`${name} returns!`) } else { console.log(`Just ${name} this time.`) } } catch (e) { if (e instanceof ReferenceError) { printError(e, true); } else { printError(e, false); } }

Sure enough, rather than giving us an output, the JavaScript engine produces the expected Invalid Left-Hand Assignment error:

It's worth noting that catching an Invalid Left-Hand Assignment error with a typical try-catch block is particular difficult, because the engine parses the code from inside out, meaning inner code blocks are parsed and executed before outer blocks. Since the issue of using a = assignment operator instead of a == comparison operator means the actual structure of the code is changed from the expected, the outer try-catch fails to be parsed and properly executed. In short, this means Invalid Left-Hand Assignment errors are always "raw", without any simple means of catching them.

Another common method for producing an Invalid Left-Hand Assignment error is when attempting to concatenate a string value onto a variable using the addition assignment += operator, instead of the concatenation operator +. For example, below we're attempting to perform concatenation on the name variable on multiple lines, but we've accidentally used the += operator:

try { var name = 'Bob' += ' Smith';

console.log(`Name is ${name}.`); } catch (e) { if (e instanceof ReferenceError) { printError(e, true); } else { printError(e, false); } }

This isn't the syntax JavaScript expects when concatenating multiple values onto a string, so an Invalid Left-Hand Assignment error is thrown:

To resolve this, we simply need to replace += with the concatenation operator +:

try { var name = 'Bob' + ' Smith';

Now we skip the Invalid Left-Hand Assignment error entirely and get our expected output indicating the full name stored in the name variable:

To dive even deeper into understanding how your applications deal with JavaScript Errors, check out the revolutionary Airbrake JavaScript error tracking tool for real-time alerts and instantaneous insight into what went wrong with your JavaScript code.

Written By: Frances Banks

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Dec 28, 2016 8:00:56 AM | JavaScript Error Handling - ReferenceError: assignment to undeclared variable “x”

Feb 15, 2017 7:41:35 am | javascript error handling: syntaxerror: "use strict" not allowed in function with non-simple parameters, may 21, 2017 9:00:51 am | javascript errors - syntaxerror: test for equality mistyped as assignment.

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• ReferenceError: invalid assignment left-hand side

ReferenceError .

What went wrong?

There was an unexpected assignment somewhere. This might be due to a mismatch of a assignment operator and a comparison operator , for example. While a single " = " sign assigns a value to a variable, the " == " or " === " operators compare a value.

In the if statement, you want to use a comparison operator ("=="), and for the string concatenation, the plus ("+") operator is needed.

• Assignment operators
• Comparison operators

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• ReferenceError: invalid assignment left-hand side

The JavaScript exception "invalid assignment left-hand side" occurs when there was an unexpected assignment somewhere. For example, a single " = " sign was used instead of " == " or " === ".

ReferenceError .

What went wrong?

There was an unexpected assignment somewhere. This might be due to a mismatch of a assignment operator and a comparison operator , for example. While a single " = " sign assigns a value to a variable, the " == " or " === " operators compare a value.

Typical invalid assignments

In the if statement, you want to use a comparison operator ("=="), and for the string concatenation, the plus ("+") operator is needed.

• Assignment operators
• Comparison operators
• JavaScript basics
• JavaScript first steps
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• Introducing JavaScript objects
• Introduction
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• Loops and iteration
• Expressions and operators
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• Details of the object model
• Using promises
• Iterators and generators
• Meta programming
• JavaScript modules
• Client-side web APIs
• A re-introduction to JavaScript
• JavaScript data structures
• Equality comparisons and sameness
• Inheritance and the prototype chain
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JavaScript ReferenceError – Invalid assignment left-hand side

This JavaScript exception invalid assignment left-hand side occurs if there is a wrong assignment somewhere in code. A single “=” sign instead of “==” or “===” is an Invalid assignment.

Error Type:

Cause of the error: There may be a misunderstanding between the assignment operator and a comparison operator.

Basic Example of ReferenceError – Invalid assignment left-hand side, run the code and check the console

Example 1: In this example, “=” operator is misused as “==”, So the error occurred.

Example 2: In this example, the + operator is used with the declaration, So the error has not occurred.

Output: 

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ReferenceError: invalid assignment left-hand side

The JavaScript exception "invalid assignment left-hand side" occurs when there was an unexpected assignment somewhere. For example, a single " = " sign was used instead of " == " or " === ".

ReferenceError .

What went wrong?

There was an unexpected assignment somewhere. This might be due to a mismatch of a assignment operator and an equality operator , for example. While a single " = " sign assigns a value to a variable, the " == " or " === " operators compare a value.

Typical invalid assignments

In the if statement, you want to use an equality operator ("=="), and for the string concatenation, the plus ("+") operator is needed.

• Assignment operators
• Equality operators

© 2005–2021 MDN contributors. Licensed under the Creative Commons Attribution-ShareAlike License v2.5 or later. https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Errors/Invalid_assignment_left-hand_side

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Syntax Error: "Invalid Left-hand Side in assignment."

I am unsure what else is necessary in my code. My syntax is as follows:

var isEven = function(number) {

if (number % 2 === 0){ return true; }

else if (isNan(number) = true) { return “This is not a number!”; }

else { return false; }

isEven (“r”)

First, I would like to know what this error code means. Secondly, I would like further insight into how one would use isNaN in this code.

Answer 5605035486f55253a300022e

if you want to compare two values you need to use == or ===, = just tries to assign the value on the left to the variable on the right. And here is the problem as you have a value on the left and not a variable which is why you get that:

To get rid of it just fix the comparison:

or get rid of the == true or === true and just use:

as a comparision with true is always what it was before e.g.

if isNan(number) is false:

and if it is true:

Answer 560a184286f552cb7c000371

if(number%2===0) { return true; } else if (isNaN(number)===true) { return “Your input type is notcorrect”; }

else { return false; } };isEven(“hi”);

worked for me

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Invalid left-hand side in assignment expression

Hello. I am attempting to create a self-generating biology question that randomly generates three numbers for the problem question, then asks a yes or no question. When I was attempting to create the function that checks for the answer to the question and compared it to the student input, I get the “Invalid left-hand side in assignment expression”

My code is here, line 33 in the JavaScript window: https://codepen.io/KDalang/pen/OJpEdQB

Here is the specific line in question: if (chiTotal <= 3.841 && input=“Yes”) What did I do wrong?

= is assignment of a value to a variable == is weak comparison (with type coercion) === is strong comparison (probably what you want)

Hey thanks for the quick reply! I actually want it to be a “less than or equal to” and I used <=. <== and <=== don’t do anything either.

Edit: Nevermind, I understand now.

Do you try to compare values or do you try to assign a value?

Oh my gosh! Sorry its 2a.m. over here I understand what you and JeremyLT are saying now. Thanks so much!

This topic was automatically closed 182 days after the last reply. New replies are no longer allowed.

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Angular error- "The left-hand side of an assignment expression may not be an optional property access.ts"?

In my Angular typescript file I have following code. I need help to solve this typescript error

I am getting the following error

I have created custom pipe where if I get loop through text and search for a word, if word is present in given text I add highlighted class to it so that, I can hilight that word with pink color

• runtime-error

3 Answers 3

Just remove the question mark in

for the error.

It should be

Because that ? making document.querySelector('.highlighted') as optional and in the right we are assigning value to that.

But it will set the pink color in the first item that has the class 'highlighted'. So you need to loop through the elements or add a pink-color class to it. And set the background color of that class item to pink

Try the below code for changing all elements with the class highlighted.

Ckeck this https://stackblitz.com/edit/angular-services-example-ygmmaq?file=src/app/app.component.html

• 3 On removing ? from document.querySelector<HTMLElement>('.highlighted').style.backgroundColor = 'white' ` it says Object is possibly 'null'.ts(2531) –  Paritosh M Commented Feb 3, 2021 at 18:40
• Try the last piece of code. That will work. It has the check, like whether it exists or not. –  Jobelle Commented Feb 5, 2021 at 4:06
• I tried above code at forEach((item:HTMLElement) it says Argument of type '(item: HTMLElement) => void' is not assignable to parameter of type '(value: Element, key: number, parent: NodeListOf<Element>) => void'. Types of parameters 'item' and 'value' are incompatible. Type 'Element' is missing the following properties from type 'HTMLElement': accessKey, accessKeyLabel, autocapitalize, dir, and 107 more. –  Paritosh M Commented Feb 6, 2021 at 0:50
• @ParitoshM I have updated the answer with stackblits link. Please check that :) –  Jobelle Commented Feb 10, 2021 at 3:42
• @ParitoshM store the result of the querySelector in a variable first. –  Henridv Commented Nov 15, 2021 at 21:05

This may help others for late reply.

The casting was not set right on the document object

Refer to this link for more info Typescript 3.7@beta Optional Chaining operator using problem

• I am confused can you take whole ngAfterViewInit() code and then add your code? –  Paritosh M Commented Feb 3, 2021 at 3:45
• @ParitoshM removing the lifecycle hook doesn't change the static type checking error. –  Richie50 Commented Feb 3, 2021 at 3:51
• Okay but what should I do in place where I have just this single line defined in other function? document.querySelector<HTMLElement>('.highlighted')?.style.backgroundColor = 'white'; –  Paritosh M Commented Feb 3, 2021 at 3:57
• I changed the answer. You have to assign the querySelector to a variable. –  Richie50 Commented Feb 3, 2021 at 3:59
• If I write as above <HTMLElement>document.querySelector('.highlighted')?.style.backgroundColor = 'white'; ` it says Property 'style' does not exist on type 'Element'.ts(2339) –  Paritosh M Commented Feb 3, 2021 at 4:00

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