Home  >  Article  >  Web Front-end  >  Convert jQuery objects and DOM objects to each other_jquery

Convert jQuery objects and DOM objects to each other_jquery

WBOY
WBOYOriginal
2016-05-16 18:53:541149browse

DOM objects are some object operations inherent in Javascript. DOM objects can use methods inherent in Javascript, but cannot use methods in jQuery. For example: document.getElementById("img").src="test.jpg"; document.getElementById("img") here is the DOM object;
$("#img").attr("src", "test.jpg"); and document.getElementById("img").src="test.jpg"; are equivalent and correct, but $("#img").src ="test.jpg"; or document.getElementById("img").attr("src","test.jpg"); are both wrong.

Let’s talk about an example: this is this. When I write jQuery, I often write like this: this.attr("src","test.jpg"); But it’s an error. In fact, this is a DOM object, and .attr("src","test.jpg") is a jQuery method, so something went wrong. To solve this problem, you need to convert the DOM object into a jQuery object, such as $(this).attr("src","test.jpg");

1. Convert DOM object to jQuery object:

For a DOM object, you only need to wrap the DOM object with $() to get a jQuery object. $(DOM object)

For example: var v=document.getElementById("v"); //DOM object

var $v=$(v); //jQuery object

After conversion, you can use jQuery methods at will.

2. Convert jQuery object to DOM object:

Two conversion methods convert a jQuery object into a DOM object: [index] and .get(index);

(1) The jQuery object is a data object, which can be used through the [index] method, to get the corresponding DOM object.

For example: var $v =$("#v"); //jQuery object

var v=$v[0]; //DOM object

alert (v.checked) //Detect whether this checkbox is selected

(2) jQuery itself provides, through the .get(index) method, the corresponding DOM object

is obtained, such as: var $v =$("#v"); //jQuery object

var v=$v.get(0); //DOM object

alert(v.checked) //Detect this Whether the checkbox is selected

Statement:
The content of this article is voluntarily contributed by netizens, and the copyright belongs to the original author. This site does not assume corresponding legal responsibility. If you find any content suspected of plagiarism or infringement, please contact admin@php.cn