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I am learning the basics of JQuery, and can't solve this problem: given 3 green <li> elements turn 1-st and 3-rd elements to red color, and the 2-nd element to orange.
Code:
<!DOCTYPE html>
<html>
<head>
<title>element</title>
<script src=".11.0/jquery.min.js" ></script>
<meta http-equiv="Content-Type" content="text/html;charset=utf-8" >
<style type="text/css" media="screen">
ul li{color: green;}
</style>
</head>
<body>
<ul>
<li>text 1</li>
<li>text 2</li>
<li>text 3</li>
</ul>
<script>
var lis = $("ul li").css("color", "red");
</script>
</body>
</html>
I was able to make all the elements red, but I can't make the 2-nd orange: lis[1].css("color", "orange"); doesn't work.
I am learning the basics of JQuery, and can't solve this problem: given 3 green <li> elements turn 1-st and 3-rd elements to red color, and the 2-nd element to orange.
Code:
<!DOCTYPE html>
<html>
<head>
<title>element</title>
<script src="http://ajax.googleapis./ajax/libs/jquery/1.11.0/jquery.min.js" ></script>
<meta http-equiv="Content-Type" content="text/html;charset=utf-8" >
<style type="text/css" media="screen">
ul li{color: green;}
</style>
</head>
<body>
<ul>
<li>text 1</li>
<li>text 2</li>
<li>text 3</li>
</ul>
<script>
var lis = $("ul li").css("color", "red");
</script>
</body>
</html>
I was able to make all the elements red, but I can't make the 2-nd orange: lis[1].css("color", "orange"); doesn't work.
- 1 Does this answer your question? $(element)[index].addClass(); does not work – Sebastian Simon Commented Jul 5, 2021 at 16:06
5 Answers
Reset to default 7You are calling css on DOM object instead of jQuery object as indexer [] gives you DOM object You need eq() instead of indexer
Live Demo
lis.eq(1).css("color", "orange");
Description: Reduce the set of matched elements to the one at the specified index.
You can also use :eq() directly in the selector
$("ul li:eq(1)").css("color", "red");
You can achieve it with pure CSS by applying :nth-child selector:
ul li:nth-child(2) {
color: red;
}
Fiddle Demo
Since you are learning jQuery, you can use the :even selector:
var lis = $('ul li');
lis.filter(':even').css('color', 'red'); // Zero based indexing selects 0 and 2
lis.filter(':odd').css('color', 'orange'); // Selects 1
Note, from the docs:
Because :even is a jQuery extension and not part of the CSS specification, queries using :even cannot take advantage of the performance boost provided by the native DOM querySelectorAll() method. To achieve the best performance when using :even to select elements, first select the elements using a pure CSS selector, then use .filter(":even").
please write your code in document.ready() and use eq
for all element
$(document).ready(function(){
$("ul li").css("color", "red");
});
for particluar element
$(document).ready(function(){
$("ul li:eq(0)").css("color", "red"); //for first element
$("ul li:eq(1)").css("color", "red");//for second element
$("ul li:eq(2)").css("color", "red");//for third element
});
If you want to select only first element than use this...
use CSS pesudo selector :first-of-type
$("li:first-of-type").css("color","orange");
or you can use jQuery built in selector
$("li:first").css("color","orange");
Both will work fine...but the jQuery method is relatively slower than CSS pesudo selector so use the first one for better performance
now if you want to select any other index then use .eq()
$(selectorName.eq(index)).css(...);
本文标签： javascriptHow to apply style to a specific element in JQuery arrayStack Overflow
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