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JS Tutorial

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JS Versions

JS Versions JS 2009 (ES5) JS 2015 (ES6) JS 2016 JS 2017 JS 2018 JS 2019 JS 2020 JS 2021 JS 2022 JS 2023 JS 2024 JS IE / Edge JS History

JS Objects

Object Definitions Object Prototypes Object Methods Object Properties Object Get / Set Object Protection

JS Functions

Function Definitions Function Parameters Function Invocation Function Call Function Apply Function Bind Function Closures

JS Classes

Class Intro Class Inheritance Class Static

JS Async

JS Callbacks JS Asynchronous JS Promises JS Async/Await

JS HTML DOM

DOM Intro DOM Methods DOM Document DOM Elements DOM HTML DOM Forms DOM CSS DOM Animations DOM Events DOM Event Listener DOM Navigation DOM Nodes DOM Collections DOM Node Lists

JS Browser BOM

JS Window JS Screen JS Location JS History JS Navigator JS Popup Alert JS Timing JS Cookies

JS Web APIs

Web API Intro Web Forms API Web History API Web Storage API Web Worker API Web Fetch API Web Geolocation API

JS AJAX

AJAX Intro AJAX XMLHttp AJAX Request AJAX Response AJAX XML File AJAX PHP AJAX ASP AJAX Database AJAX Applications AJAX Examples

JS JSON

JSON Intro JSON Syntax JSON vs XML JSON Data Types JSON Parse JSON Stringify JSON Objects JSON Arrays JSON Server JSON PHP JSON HTML JSON JSONP

JS vs jQuery

jQuery Selectors jQuery HTML jQuery CSS jQuery DOM

JS Graphics

JS Graphics JS Canvas JS Plotly JS Chart.js JS Google Chart JS D3.js

JS Examples

JS Examples JS HTML DOM JS HTML Input JS HTML Objects JS HTML Events JS Browser JS Editor JS Exercises JS Quiz JS Website JS Syllabus JS Study Plan JS Interview Prep JS Bootcamp JS Certificate

JS References

JavaScript Objects HTML DOM Objects


JavaScript Common Mistakes


This chapter points out some common JavaScript mistakes.


Accidentally Using the Assignment Operator

JavaScript programs may generate unexpected results if a programmer accidentally uses an assignment operator (=), instead of a comparison operator (==) in an if statement.

This if statement returns false (as expected) because x is not equal to 10:

let x = 0;
if (x == 10)
Try it Yourself »

This if statement returns true (maybe not as expected), because 10 is true:

let x = 0;
if (x = 10)
Try it Yourself »

This if statement returns false (maybe not as expected), because 0 is false:

let x = 0;
if (x = 0)
Try it Yourself »

An assignment always returns the value of the assignment.


Expecting Loose Comparison

In regular comparison, data type does not matter. This if statement returns true:

let x = 10;
let y = "10";
if (x == y)
Try it Yourself »

In strict comparison, data type does matter. This if statement returns false:

let x = 10;
let y = "10";
if (x === y)
Try it Yourself »

It is a common mistake to forget that switch statements use strict comparison:

This case switch will display an alert:

let x = 10;
switch(x) {
  case 10: alert("Hello");
}
Try it Yourself »

This case switch will not display an alert:

let x = 10;
switch(x) {
  case "10": alert("Hello");
}
Try it Yourself »


Confusing Addition & Concatenation

Addition is about adding numbers.

Concatenation is about adding strings.

In JavaScript both operations use the same + operator.

Because of this, adding a number as a number will produce a different result from adding a number as a string:

let x = 10;
x = 10 + 5;       // Now x is 15

let y = 10;
y += "5";        // Now y is "105"
Try it Yourself »

When adding two variables, it can be difficult to anticipate the result:

let x = 10;
let y = 5;
let z = x + y;     // Now z is 15

let x = 10;
let y = "5";
let z = x + y;     // Now z is "105"
Try it Yourself »

Misunderstanding Floats

All numbers in JavaScript are stored as 64-bits Floating point numbers (Floats).

All programming languages, including JavaScript, have difficulties with precise floating point values:

let x = 0.1;
let y = 0.2;
let z = x + y            // the result in z will not be 0.3
Try it Yourself »

To solve the problem above, it helps to multiply and divide:

Example

let z = (x * 10 + y * 10) / 10;       // z will be 0.3
Try it Yourself »

Breaking a JavaScript String

JavaScript will allow you to break a statement into two lines:

Example 1

let x =
"Hello World!";
Try it Yourself »

But, breaking a statement in the middle of a string will not work:

Example 2

let x = "Hello
World!";
Try it Yourself »

You must use a "backslash" if you must break a statement in a string:

Example 3

let x = "Hello \
World!";
Try it Yourself »

Misplacing Semicolon

Because of a misplaced semicolon, this code block will execute regardless of the value of x:

if (x == 19);
{
  // code block 
}
Try it Yourself »

Breaking a Return Statement

It is a default JavaScript behavior to close a statement automatically at the end of a line.

Because of this, these two examples will return the same result:

Example 1

function myFunction(a) {
  let power = 10 
  return a * power
}
Try it Yourself »

Example 2

function myFunction(a) {
  let power = 10;
  return a * power;
}
Try it Yourself »

JavaScript will also allow you to break a statement into two lines.

Because of this, example 3 will also return the same result:

Example 3

function myFunction(a) {
  let
  power = 10; 
  return a * power;
}
Try it Yourself »

But, what will happen if you break the return statement in two lines like this:

Example 4

function myFunction(a) {
  let
  power = 10; 
  return
  a * power;
}
Try it Yourself »

The function will return undefined!

Why? Because JavaScript thought you meant:

Example 5

function myFunction(a) {
  let
  power = 10; 
  return;
  a * power;
}
Try it Yourself »

Explanation

If a statement is incomplete like:

let

JavaScript will try to complete the statement by reading the next line:

power = 10;

But since this statement is complete:

return

JavaScript will automatically close it like this:

return;

This happens because closing (ending) statements with semicolon is optional in JavaScript.

JavaScript will close the return statement at the end of the line, because it is a complete statement.

Never break a return statement.


Accessing Arrays with Named Indexes

Many programming languages support arrays with named indexes.

Arrays with named indexes are called associative arrays (or hashes).

JavaScript does not support arrays with named indexes.

In JavaScript, arrays use numbered indexes:  

Example

const person = [];
person[0] = "John";
person[1] = "Doe";
person[2] = 46;
person.length;       // person.length will return 3
person[0];           // person[0] will return "John"
Try it Yourself »

In JavaScript, objects use named indexes.

If you use a named index, when accessing an array, JavaScript will redefine the array to a standard object.

After the automatic redefinition, array methods and properties will produce undefined or incorrect results:

Example:

const person = [];
person["firstName"] = "John";
person["lastName"] = "Doe";
person["age"] = 46;
person.length;      // person.length will return 0
person[0];          // person[0] will return undefined
Try it Yourself »

Ending Definitions with a Comma

Trailing commas in object and array definition are legal in ECMAScript 5.

Object Example:

person = {firstName:"John", lastName:"Doe", age:46,}

Array Example:

points = [40, 100, 1, 5, 25, 10,];

WARNING !!

Internet Explorer 8 will crash.

JSON does not allow trailing commas.

JSON:

person = {"firstName":"John", "lastName":"Doe", "age":46}

JSON:

points = [40, 100, 1, 5, 25, 10];

Undefined is Not Null

JavaScript objects, variables, properties, and methods can be undefined.

In addition, empty JavaScript objects can have the value null.

This can make it a little bit difficult to test if an object is empty.

You can test if an object exists by testing if the type is undefined:

Example:

if (typeof myObj === "undefined") 
Try it Yourself »

But you cannot test if an object is null, because this will throw an error if the object is undefined:

Incorrect:

if (myObj === null) 

To solve this problem, you must test if an object is not null, and not undefined.

But this can still throw an error:

Incorrect:

if (myObj !== null && typeof myObj !== "undefined") 

Because of this, you must test for not undefined before you can test for not null:

Correct:

if (typeof myObj !== "undefined" && myObj !== null) 
Try it Yourself »


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