What do querySelectorAll and getElementsBy* methods return?

问题

Do getElementsByClassName (and similar functions like getElementsByTagName and querySelectorAll) work the same as getElementById or do they return an array of elements?

The reason I ask is because I am trying to change the style of all elements using getElementsByClassName. See below.

//doesn\'t work
document.getElementsByClassName(\'myElement\').style.size = \'100px\';

//works
document.getElementById(\'myIdElement\').style.size = \'100px\';

回答1:

Your getElementById() code works since IDs have to be unique and thus the function always returns exactly one element (or null if none was found).

However, getElementsByClassName(), querySelectorAll(), and other getElementsBy* methods return an array-like collection of elements. Iterate over it like you would with a real array:

var elems = document.getElementsByClassName('myElement');
for(var i = 0; i < elems.length; i++) {
    elems[i].style.size = '100px';
}

If you prefer something shorter, consider using jQuery:

$('.myElement').css('size', '100px');

回答2:

You are using a array as an object, the difference between getElementbyId and getElementsByClassName is that:

getElementbyId will return an Element object or null if no element with the ID is found
getElementsByClassName will return a live HTMLCollection, possibly of length 0 if no matching elements are found

getElementsByClassName

The getElementsByClassName(classNames) method takes a string that contains an unordered set of unique space-separated tokens representing classes. When called, the method must return a live NodeList object containing all the elements in the document that have all the classes specified in that argument, having obtained the classes by splitting a string on spaces. If there are no tokens specified in the argument, then the method must return an empty NodeList.

https://www.w3.org/TR/2008/WD-html5-20080610/dom.html#getelementsbyclassname

getElementById

The getElementById() method accesses the first element with the specified id.

https://developer.mozilla.org/en-US/docs/Web/API/Document/getElementById

in your code the lines:

1- document.getElementsByClassName('myElement').style.size = '100px';

will NOT work as expected, because the getElementByClassName will return an array, and the array will NOT have the style property, you can access each element by iterating through them.

That's why the function getElementById worked for you, this function will return the direct object. Therefore you will be able to access the style property.

回答3:

The following description is taken from this page:

The getElementsByClassName() method returns a collection of all elements in the document with the specified class name, as a NodeList object.

The NodeList object represents a collection of nodes. The nodes can be accessed by index numbers. The index starts at 0.

Tip: You can use the length property of the NodeList object to determine the number of elements with a specified class name, then you can loop through all elements and extract the info you want.

So, as a parameter getElementsByClassName would accept a class name.

If this is your HTML body:

<div id="first" class="menuItem"></div>
<div id="second" class="menuItem"></div>
<div id="third" class="menuItem"></div>
<div id="footer"></div>

then var menuItems = document.getElementsByClassName('menuItem') would return a collection (not an array) of the 3 upper <div>s, as they match the given class name.

You can then iterate over this nodes (<div>s in this case) collection with:

for (var menuItemIndex = 0 ; menuItems.length ; menuItemIndex ++) {
   var currentMenuItem = menuItems[menuItemIndex];
   // do stuff with currentMenuItem as a node.
}

Please refer to this post for more on differences between elements and nodes.

回答4:

ES6 provides Array.from() method, which creates a new Array instance from an array-like or iterable object.

let boxes = document.getElementsByClassName('box');

Array.from(boxes).forEach(v => v.style.background = 'green');
console.log(Array.from(boxes));

.box {
  width: 50px;
  height: 50px;
  margin: 5px;
  background: blue;
  display: inline-block;
}

<div class='box'></div>
<div class='box'></div>
<div class='box'></div>
<div class='box'></div>

As you can see inside the code snippet, after using Array.from() function you are then able to manipulate over each element.

The same solution using jQuery.

$('.box').css({'background':'green'});

.box {
  width: 50px;
  height: 50px;
  margin: 5px;
  background: blue;
  display: inline-block;
}

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div class='box'></div>
<div class='box'></div>
<div class='box'></div>
<div class='box'></div>

回答5:

In Other Words

document.querySelector() selects only the first one element of the specified selector. So it doesn't spit out an array, it's a single value. Similar to document.getElementById() which fetches ID-elements only, since IDs have to be unique.
document.querySelectorAll() selects all elements with the specified selector and returns them in an array. Similar to document.getElementsByClassName() for classes and document.getElementsByTagName() tags only.

Why use querySelector?

It's used merely for the sole purpose of ease and brevity.

Why use getElement/sBy?*

Faster performance.

Why this performance difference?

Both ways of selection has the purpose of creating a NodeList for further use. querySelectors generates a static NodeList with the selectors thus it must be first created from scratch.
getElement/sBy* immediately adapts the existing live NodeList of the current DOM.

So, when to use which method it's up to you/your project/your device.

Infos

Demo of all methods
NodeList Documentation
Performance Test

回答6:

It returns Array-like list.

You make that an Array as example

var el = getElementsByClassName("elem");
el = Array.prototype.slice.call(el); //this line
el[0].appendChild(otherElem);

回答7:

You could get a single element by running

document.querySelector('.myElement').style.size = '100px';

but it's going to work for the first element with class .myElement.

If you would like apply this for all elements with the class I suggest you to use

document.querySelectorAll('.myElement').forEach(function(element) {
    element.style.size = '100px';
});

回答8:

/*
 * To hide all elements with the same class, 
 * use looping to reach each element with that class. 
 * In this case, looping is done recursively
 */

const hideAll = (className, i=0) => {
if(!document.getElementsByClassName(className)[i]){ //exits the loop when element of that id does not exist
  return; 
}

document.getElementsByClassName(className)[i].style.visibility = 'hidden'; //hide element
return hideAll(className, i+1) //loop for the next element
}

hideAll('appBanner') //the function call requires the class name

回答9:

With ES5+ (any browsed nowadays - 2017) you should be able to do

[].forEach.call(document.getElementsByClassName('answer'), function(el) {
    el.style.color= 'red';
});

来源：https://stackoverflow.com/questions/10693845/what-do-queryselectorall-and-getelementsby-methods-return

标签

javascript

getelementsbyclassname

dom-traversal