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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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How to fix SyntaxError: invalid assignment left-hand side

by Nathan Sebhastian

Posted on Jul 10, 2023

Reading time: 3 minutes

invalid left hand side in assignment expression innerhtml

When running JavaScript code, you might encounter an error that says:

Both errors are the same, and they occured when you use the single equal = sign instead of double == or triple === equals when writing a conditional statement with multiple conditions.

Let me show you an example that causes this error and how I fix it.

How to reproduce this error

Suppose you have an if statement with two conditions that use the logical OR || operator.

You proceed to write the statement as follows:

When you run the code above, you’ll get the error:

This error occurs because you used the assignment operator with the logical OR operator.

An assignment operator doesn’t return anything ( undefined ), so using it in a logical expression is a wrong syntax.

How to fix this error

To fix this error, you need to replace the single equal = operator with the double == or triple === equals.

Here’s an example:

By replacing the assignment operator with the comparison operator, the code now runs without any error.

The double equal is used to perform loose comparison, while the triple equal performs a strict comparison. You should always use the strict comparison operator to avoid bugs in your code.

Other causes for this error

There are other kinds of code that causes this error, but the root cause is always the same: you used a single equal = when you should be using a double or triple equals.

For example, you might use the addition assignment += operator when concatenating a string:

The code above is wrong. You should use the + operator without the = operator:

Another common cause is that you assign a value to another value:

This is wrong because you can’t assign a value to another value.

You need to declare a variable using either let or const keyword, and you don’t need to wrap the variable name in quotations:

You can also see this error when you use optional chaining as the assignment target.

For example, suppose you want to add a property to an object only when the object is defined:

Here, we want to assign the age property to the person object only when the person object is defined.

But this will cause the invalid assignment left-hand side error. You need to use the old if statement to fix this:

Now the error is resolved.

The JavaScript error SyntaxError: invalid assignment left-hand side occurs when you have an invalid syntax on the left-hand side of the assignment operator.

This error usually occurs because you used the assignment operator = when you should be using comparison operators == or === .

Once you changed the operator, the error would be fixed.

I hope this tutorial helps. Happy coding!

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

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Last updated: Jan 23, 2023 Reading time · 4 min

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# 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.

left hand side of assignment expression may not be optional property

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.

use if statement as type guard to solve the error

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.

using non null assertion to solve the error

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.

avoiding the error with type assertion

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.

using logical and operator to get around 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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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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How to solve “Invalid left-hand side in assignment” in JavaScript

"Invalid left-hand side in assignment" in JavaScript

The “Invalid left-hand side in assignment” in JavaScript is a syntax error that often occurs in comparing values, such as using the “=” sign to compare. This article will give examples of everyday situations and how to fix them.

Table of Contents

What causes the “Invalid left-hand side in assignment” in JavaScript?

This is a very common syntax error. There are many causes of errors, such as wrong comparison signs, using “=” to compare two values, and not creating a variable to receive a value from the function. Here are examples of some errors.

Using the wrong comparison sign

“Invalid left-hand side in assignment” is an error caused by misspelled operator when comparing two values.

Using “=” to compare two values

This error also happens because instead of using “===”, you use “=” to compare.

Not creating a variable to receive a value from the function

This case leaves the wrong position of the variable.

Do not use square brackets when accessing object properties

For properties like this, we need to use square brackets.

Solution for the error “Invalid left-hand side in assignment” in JavaScript

Use the correct operator.

We need to pay attention to the comparison signs in expressions.

Pay attention to the position on either side of the “=”

To get the value of a function, we need to create a variable to the left of the “=” sign.

Use square brackets when accessing properties

For properties of objects with two or more words, we use square brackets.

The article has given some examples that lead to the error “Invalid left-hand side in assignment” in JavaScript. These are just syntax errors , pay attention to the process of writing code, and we will avoid such errors. We hope you can fix it quickly through this article. Good luck to you!

Maybe you are interested :

  • TypeError: Assignment to Constant Variable in JavaScript
  • Element type is invalid, expected a string (for built in components) or a class/function but got – How to solve?
  • RangeError: Invalid time value in JavaScript

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Carolyn Hise has three years of software development expertise. Strong familiarity with the following languages is required: Python, Typescript/Nodejs, .Net, Java, C++, and a strong foundation in Object-oriented programming (OOP).

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  • Destructuring assignment

Unpacking values from a regular expression match

Es2015 version, invalid javascript identifier as a property name.

The destructuring assignment syntax is a JavaScript expression that makes it possible to unpack values from arrays, or properties from objects, into distinct variables.

Description

The object and array literal expressions provide an easy way to create ad hoc packages of data.

The destructuring assignment uses similar syntax, but on the left-hand side of the assignment to define what values to unpack from the sourced variable.

This capability is similar to features present in languages such as Perl and Python.

Array destructuring

Basic variable assignment, assignment separate from declaration.

A variable can be assigned its value via destructuring separate from the variable's declaration.

Default values

A variable can be assigned a default, in the case that the value unpacked from the array is undefined .

Swapping variables

Two variables values can be swapped in one destructuring expression.

Without destructuring assignment, swapping two values requires a temporary variable (or, in some low-level languages, the XOR-swap trick ).

Parsing an array returned from a function

It's always been possible to return an array from a function. Destructuring can make working with an array return value more concise.

In this example, f() returns the values [1, 2] as its output, which can be parsed in a single line with destructuring.

Ignoring some returned values

You can ignore return values that you're not interested in:

You can also ignore all returned values:

Assigning the rest of an array to a variable

When destructuring an array, you can unpack and assign the remaining part of it to a variable using the rest pattern:

Note that a SyntaxError will be thrown if a trailing comma is used on the left-hand side with a rest element:

When the regular expression exec() method finds a match, it returns an array containing first the entire matched portion of the string and then the portions of the string that matched each parenthesized group in the regular expression. Destructuring assignment allows you to unpack the parts out of this array easily, ignoring the full match if it is not needed.

Object destructuring

Basic assignment, assignment without declaration.

A variable can be assigned its value with destructuring separate from its declaration.

The ( .. ) around the assignment statement is required syntax when using object literal destructuring assignment without a declaration.

{a, b} = {a: 1, b: 2} is not valid stand-alone syntax, as the {a, b} on the left-hand side is considered a block and not an object literal.

However, ({a, b} = {a: 1, b: 2}) is valid, as is var {a, b} = {a: 1, b: 2}

NOTE: Your ( ..) expression needs to be preceded by a semicolon or it may be used to execute a function on the previous line.

Assigning to new variable names

A property can be unpacked from an object and assigned to a variable with a different name than the object property.

A variable can be assigned a default, in the case that the value unpacked from the object is undefined .

Setting a function parameter's default value

Es5 version, nested object and array destructuring, for of iteration and destructuring, unpacking fields from objects passed as function parameter.

This unpacks the id , displayName and firstName from the user object and prints them.

Computed object property names and destructuring

Computed property names, like on object literals , can be used with destructuring.

Rest in Object Destructuring

The Rest/Spread Properties for ECMAScript proposal (stage 3) adds the rest syntax to destructuring. Rest properties collect the remaining own enumerable property keys that are not already picked off by the destructuring pattern.

Destructuring can be used with property names that are not valid JavaScript identifiers  by providing an alternative identifer that is valid.

Specifications

Browser compatibility.

[1] Requires "Enable experimental Javascript features" to be enabled under `about:flags`

Firefox-specific notes

  • Firefox provided a non-standard language extension in JS1.7 for destructuring. This extension has been removed in Gecko 40 (Firefox 40 / Thunderbird 40 / SeaMonkey 2.37). See bug 1083498 .
  • Starting with Gecko 41 (Firefox 41 / Thunderbird 41 / SeaMonkey 2.38) and to comply with the ES2015 specification, parenthesized destructuring patterns, like ([a, b]) = [1, 2] or ({a, b}) = { a: 1, b: 2 } , are now considered invalid and will throw a SyntaxError . See Jeff Walden's blog post and bug 1146136 for more details.
  • Assignment operators
  • "ES6 in Depth: Destructuring" on hacks.mozilla.org

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Learn Form Validation by Building a Calorie Counter - Step 52

I have no clue wtf does it want me to do!

It says as follows: " Your other bug occurs if you add a Breakfast entry, fill it in, then add a second Breakfast entry. You’ll see that the values you added disappeared.

This is because you are updating innerHTML directly, which does not preserve your input content. Change your innerHTML assignment to use the insertAdjacentHTML() method of targetInputContainer instead. Do not pass any arguments yet."

And I’ve done it, I changed “.innerHTML” to “.insertAdjacentHTML()” and I doesn’t fucking work

My code so far:

Challenge information:.

Hi there and welcome to our community!

Change your innerHTML assignment to use the insertAdjacentHTML() method of targetInputContainer instead. Do not pass any arguments yet.

You don’t concatenate using += when using the .insertAdjacentHTML() method. Instead, you will pass arguments inside the parentheses. For now, though just add the method, with empty parentheses (i.e. remove += HTMLString ).

Thanks, bro. I’m just dumb then lol… I just could not see it, my bad.

I understand that passing arguments to the insertAdjacentHTML() removes the need to concatenate to the string (only after reading this post), but if we’re just learning a new method, perhaps this should have been noted in the step? Not criticizing and I’m sure you’re not responsible for the wording of this step, but just wanted to put that thought out there if you have access to those who are responsible for making edits (I know this is beta ;-)). The only information we noobs get from this step is we’re not supposed to pass arguments yet.

A note to all the noobs like me, when nothing seems to happen when you try to check your code, check the “Console” tab in the “Preview” panel on the right, and it should give you some error messages that could give you some sort of lead on what the issue is. I always forget about that tab lol. For example, when I checked my code before making the corrections for this step, I received the following:

SyntaxError: unknown: Invalid left-hand side in assignment expression. (32:2) 30 | placeholder=“Calories” 31 | />`; 32 | targetInputContainer.insertAdjacentHTML() += HTMLString; | ^ 33 | } 34 | 35 | addEntryButton.addEventListener(“click”, addEntry); SyntaxError: unknown: Invalid left-hand side in assignment expression. (32:2)

30 | placeholder=“Calories” 31 | />`;

32 | targetInputContainer.insertAdjacentHTML( ) += HTMLString; | ^ 33 | } 34 | 35 | addEntryButton.addEventListener(“click”, addEntry); SyntaxError: unknown: Invalid left-hand side in assignment expression. (32:2)
32 | targetInputContainer.insertAdjacentHTML() += HTMLString; | ^ 33 | } 34 | 35 | addEntryButton.addEventListener(“click”, addEntry);

Thank you for this! I was having the same issue, and would have never thought that the += HTMLString needed to be removed.

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: bad expression: Invalid left-hand side in assignment

Dazakiwi38

[[Open door|Sorter1][$CurrentPos.LocEntry[3] = 5; $CurrentPos.LocWall[3] = 5; $Reset = true; $DoorOpened = true]]
Originally I did try that normal link code method and what i was trying to achieve wasn't working so i wrongly assumed that you couldn't have [] within the link|passage code because it uses them, i was assuming it wasn't passing the values.
// WRONG: The closing square bracket of the array touches the closing pair of the markup, making them ambiguous. [[Go where?|There][$list to ["A", "B"]]] // CORRECT: The closing square bracket of the array is separated from the closing pair of the markup by a space. [[Go where?|There][$list to ["A", "B"] ]]

質問をすることでしか得られない、回答やアドバイスがある。

15分調べてもわからないことは、質問しよう!.

JavaScriptは、プログラミング言語のひとつです。ネットスケープコミュニケーションズで開発されました。 開発当初はLiveScriptと呼ばれていましたが、業務提携していたサン・マイクロシステムズが開発したJavaが脚光を浴びていたことから、JavaScriptと改名されました。 動きのあるWebページを作ることを目的に開発されたもので、主要なWebブラウザのほとんどに搭載されています。

JavaScriptリターンされたオブジェクトに直接値を代入する

hakasensei

投稿 2016/06/12 15:22

javascriptでオブジェクトをリターンする関数を作り、それをまるまる左辺においてみたのですが、 Uncaught ReferenceError: Invalid left-hand side in assignment というエラーが出てしまいました。 なぜでしょうか?

初心者です。オブジェクトの理解が生半可です。

追記 例に出したコードが言おうとしていることと違うことに気づいたので訂正させていただきます。

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eval は、引数をJavaScript構文として評価してその値を返します。 書かれているコードがすべてだとすると、 object.number の値は undefined という値になり、それへの代入はエラーになります。 なお、 undefined = 12; となるわけじゃなくて、 (undefinedの値) = 12; となります。 値への代入という意味では 1 = 12; 等と同じエラーメッセージですね。

object の number プロパティに代入したいのなら、例えば、

追記: タイトルと質問内容が合ってない気がします。

投稿 2016/06/12 16:24

otn

総合スコア 84165

2016/06/12 22:17

2016/06/12 22:24

javascriptでオブジェクトをリターンする関数を作り

いいえ.オブジェクトをリターンするならこうなるはずです.

eval('object.number') はまだ存在していないので, undefined をリターンしていることになります.

しかし本質は, 関数の実行結果に対して代入しようとしている というところにあります.代入演算子の左側に来れるのは**「変数シンボル」 (オブジェクトのプロパティを含む)のみに限られます. 「値」**に対しての代入できません.

(注意) ここでいう「シンボル」はECMA Script 6の Symbol とは無関係です

この2つの x の違いはわかりますか?1つ目は変数シンボルであり,2つ目は式です.式は評価されると値になります.JavaScriptは変数シンボルと式を区別しませんが,例えばシェルスクリプトは $ の有無で区別しますよね.

投稿 2016/06/12 16:35

mpyw

2016/06/12 16:49

2016/06/12 22:03

2016/06/12 22:04

2016/06/12 23:34 編集

2016/06/13 09:26

こういうのを想定されているんですかね?

投稿 2016/06/13 02:02

hidekichi

2016/06/13 09:22

割り当て方の問題ではないですか?

投稿 2016/06/12 15:59

toshi_yama

2016/06/12 21:50

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语法错误:左侧赋值无效

当某处发生意外分配时,会发生 JavaScript 异常 "invalid assignment left-hand side" 。当使用单个 = 标志而不是 == 或 === 时,可能会触发它。

SyntaxError 或 ReferenceError ,具体取决于语法。

某处有一个意想不到的任务。例如,这可能是由于 assignment operator 和 equality operator 不匹配造成的。虽然单个 = 符号将值分配给变量,但 == 或 === 运算符比较值。

在 if 语句中,您想要使用相等运算符 ( === ),并且对于字符串连接,需要加号 ( + ) 运算符。

无效的赋值并不总是会产生语法错误。有时语法几乎是正确的,但在运行时,左侧表达式的计算结果是一个值而不是引用,因此赋值仍然无效。此类错误会在执行稍后(即实际执行该语句时)发生。

函数调用、 new 调用、 super() 和 this 都是值而不是引用。如果您想在左侧使用它们,则赋值目标需要是它们生成值的属性。

注意:在 Firefox 和 Safari 中,第一个示例在非严格模式下生成 ReferenceError ,在 strict mode 下生成 SyntaxError 。Chrome 为严格和非严格模式抛出运行时 ReferenceError 。

使用可选链作为分配目标

Optional chaining 不是有效的分配目标。

相反,你必须首先保护无效的情况。

  • Assignment operators
  • Equality operators

© 2005–2023 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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  1. Invalid Left Hand Side In Assignment

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  3. "Invalid left-hand side in assignment": incorrectly reported as

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  4. JavaScript ReferenceError

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  5. Invalid Left Hand Side In Assignment

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  6. How to solve "Invalid left-hand side in assignment" in JavaScript

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COMMENTS

  1. SyntaxError: invalid assignment left-hand side

    Invalid assignments don't always produce syntax errors. Sometimes the syntax is almost correct, but at runtime, the left hand side expression evaluates to a value instead of a reference, so the assignment is still invalid. Such errors occur later in execution, when the statement is actually executed. js

  2. Uncaught ReferenceError : Invalid left-hand side in assignment

    2 Answers Sorted by: 4 The following line is wrong: document.getElementById ("display") = "a+''+b+''+c+''+d+''+e+''+f+''+g+''+h"; You can't assign a string directly to an HTML element. One way to set the element's content is with .innerHTML: document.getElementById ("display").innerHTML = "something";

  3. JavaScript ReferenceError

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  4. How to fix SyntaxError: invalid assignment left-hand side

    SyntaxError: Invalid left-hand side in assignment This error occurs because you used the assignment operator with the logical OR operator. An assignment operator doesn't return anything ( undefined ), so using it in a logical expression is a wrong syntax. How to fix this error

  5. The left-hand side of assignment expression may not be an optional

    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. index.ts

  6. ReferenceError: invalid assignment left-hand side

    // ReferenceError: invalid assignment left-hand side In the if statement, you want to use a comparison operator ("=="), and for the string concatenation, the plus ("+") operator is needed. if (Math.PI == 3 || Math.PI == 4) { console.log ('no way!'); } var str = 'Hello, ' + 'from the ' + 'other side!'; See also Tags: Errors JavaScript ReferenceError

  7. Errors: 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 " === ". Message ReferenceError: invalid assignment left-hand side Error type ReferenceError. What went wrong? There was an unexpected assignment somewhere.

  8. How to solve "Invalid left-hand side in assignment" in JavaScript

    The "Invalid left-hand side in assignment" in JavaScript is a syntax error that often occurs in comparing values, such as using the "=" sign to compare. This article will give examples of everyday situations and how to fix them. What causes the "Invalid left-hand side in assignment" in JavaScript?

  9. Invalid left-hand side in assignment expression

    Invalid left-hand side in assignment expression - JavaScript - The freeCodeCamp Forum Invalid left-hand side in assignment expression JavaScript KDalang June 8, 2021, 5:43am 1 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.

  10. Destructuring assignment

    The ( ..) around the assignment statement is required syntax when using object literal destructuring assignment without a declaration. {a, b} = {a: 1, b: 2} is not valid stand-alone syntax, as the {a, b} on the left-hand side is considered a block and not an object literal. However, ({a, b} = {a: 1, b: 2}) is valid, as is var {a, b} = {a: 1, b: 2} NOTE: Your ( ..) expression needs to be ...

  11. ReferenceError: invalid assignment left-hand side

    ReferenceError: invalid assignment left-hand side. JavaScript の例外 "invalid assignment left-hand side" は、どこかで予想外の代入が行われたときに発生します。. 例えば、単一の " = " の記号が " == " や " === " の代わりに使用された場合です。.

  12. Learn Form Validation by Building a Calorie Counter

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  13. ReferenceError: invalid assignment left-hand side

    TypeError: invalid assignment to const "x". TypeError: More arguments needed. TypeError: property "x" is non-configurable and can't be deleted. TypeError: Reduce of empty array with no initial value. TypeError: setting getter-only property "x". TypeError: X.prototype.y called on incompatible type.

  14. : bad expression: Invalid left-hand side in assignment

    It means you're terminating your expression in the middle of it, causing the string concatenation operators to be seen on the left-hand side of an expression (i.e. the " + 5; + bitthe semi-colon terminates the current expression, which starts a new expression with the string concatenation operator). You're also missing some quotes and string concatenation operators.

  15. JavaScriptリターンされたオブジェクトに直接値を代入する

    javascriptでオブジェクトをリターンする関数を作り、それをまるまる左辺においてみたのですが、 Uncaught ReferenceError: Invalid left-hand side in assignment というエラーが出てしまいました。 なぜでしょうか? 初心者です。オブジェクトの理解が生半可です。

  16. JavaScript

    当某处发生意外分配时,会发生 JavaScript 异常 "invalid assignment left-hand side" 。 当使用单个 = 标志而不是 == 或 === 时,可能会触发它。 Message SyntaxError: Invalid left-hand side in assignment (V8-based) SyntaxError: invalid assignment left-hand side (Firefox) SyntaxError: Left side of assignment is not a reference. (Safari) Error type SyntaxError 或 ReferenceError ,具体取决于语法。 什么地方出了错?

  17. Invalid left-hand side in assignment Javascript

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  18. Js convert string to number

    Synta圎rror: invalid regular expression flag "x". Synta圎rror: invalid assignment left-hand side. Synta圎rror: identifier starts immediately after numeric literal. Synta圎rror: function statement requires a name. Synta圎rror: for-in loop head declarations may not have initializers. Synta圎rror: continue must be inside loop.

  19. javascript

    1 I am trying to make a to do list with a edit mode and view mode. I get this error: ReferenceError: Invalid left-hand side in assignment at toggleEditMode (/script.js:8:18) at HTMLButtonElement.onclick (/:14:54) Full can be found at https://repl.it/@UCYT5040/Notekeeper, but the Javascript is below.