jQuery 在jQuery中将所有hrefs作为数组获取
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/471606/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Get all hrefs as an array in jQuery
提问by Grimace of Despair
My code looks like this:
我的代码如下所示：
<ul id="ulList">
<li class="listClass" id="id1"><a href="http://link1">Link 1</a></li>
<li class="listClass" id="id2"><a href="http://link2">Link 2</a></li>
<li class="listClass" id="id3"><a href="http://link3">Link 3</a></li>
</ul>
Now I like to get the following:
现在我想得到以下内容：
All links as an array
所有链接作为一个数组
All ids of li as an array
li 的所有 id 作为数组
Can someone help me please?
有人能帮助我吗？
回答by airportyh
var ids = new Array();
var hrefs = new Array();
$('#ulList li').each(function(){
ids.push($(this).attr('id'));
hrefs.push($(this).find('a').attr('href'));
})
回答by Grimace of Despair
I know this is old, but as I like the oneliners that jQuery allows you to write, I thought I'd add it:
我知道这是旧的，但是因为我喜欢 jQuery 允许您编写的单行程序，所以我想我会添加它：
var allLinks = $('#ulList a').map(function(i,el) { return $(el).attr('href'); }).get();
var allIds = $('#ulList li').map(function(i,el) { return $(el).attr('id'); }).get();
回答by Venkat D.
Stumbled into this question and came up with a more reusable answer:
偶然发现了这个问题，并提出了一个更可重用的答案：
$.fn.collect = function(fn) {
var values = [];
if (typeof fn == 'string') {
var prop = fn;
fn = function() { return this.attr(prop); };
}
$(this).each(function() {
var val = fn.call($(this));
values.push(val);
});
return values;
};
var ids = $('#ulList li').collect('id');
var links = $('#ulList a').collect('href');
You can also pass a function into collect like so:
您还可以将函数传递给 collect ，如下所示：
var widths = $('#ulList li').collect(function() {
return this.width();
});
回答by nickf
This should work.
这应该有效。
var ids = [],
hrefs = []
;
$('#ulList')
.find('a[href]') // only target <a>s which have a href attribute
.each(function() {
hrefs.push(this.href);
})
.end()
.find('li[id]') // only target <li>s which have an id attribute
.each(function() {
ids.push(this.id);
})
;
// ids = ['id1', 'id2', 'id3']
// hrefs = ['http://link1', 'http://link2', 'http://link3']
回答by Mauricio Wolff
var links = [], ids = [];
var $ul = $('#ulList');
var $lis = $ul.children('li');
var $as = $lis.children('a');
for(var count = $lis.length-1, i = count; i >= 0; i--){
ids.push($lis[i].id);
links.push($as[i].href);
}
回答by Daniel Lizik
If you like one liners and hate having to instantiate an empty array.
如果你喜欢一个衬垫并且讨厌必须实例化一个空数组。
[]
.slice
.call($('#ulList a'))
.map(el => el.getAttribute('href'))
