jquery - 收集元素的顺序

Question

我正在编写用户根据需要更改跨度元素顺序的代码。我需要收集跨度元素的顺序并将其放入变量中，例如，如果顺序如下所示：

<span id="span-2">[<a class="move-up">Up</a>/<a class="move-down">Down</a>]</span>
<span id="span-3">[<a class="move-up">Up</a>/<a class="move-down">Down</a>]</span>
<span id="span-5">[<a class="move-up">Up</a>/<a class="move-down">Down</a>]</span>
<span id="span-4">[<a class="move-up">Up</a>/<a class="move-down">Down</a>]</span>
<span id="span-1">[<a class="move-up">Up</a>/<a class="move-down">Down</a>]</span>

那么，变量将是“2,3,5,4,1”

Answer 1

var ids = [];
$("span").each(function(index) {
    ids.push($(this).attr('id'));
});

你将在数组 ids 中有ids。对于您的示例，它将是ids = ["span-2", "span-3", "span-5", "span-4", "span-1"]

Answer 2

如果有可识别的父母，您可以这样做：

var order = new Array();
$("#parentID span").each(function(){
    order.push($(this).attr('id').substring(5));
});

但是，如果我尝试这个，我会添加一些classes 和data-ids ：

HTML：

<span class="mySpan" data-id="2" id="span-2">[<a class="move-up">Up</a>/<a class="move-down">Down</a>]</span>
<span class="mySpan" data-id="3" id="span-3">[<a class="move-up">Up</a>/<a class="move-down">Down</a>]</span>
<span class="mySpan" data-id="5" id="span-5">[<a class="move-up">Up</a>/<a class="move-down">Down</a>]</span>
<span class="mySpan" data-id="4" id="span-4">[<a class="move-up">Up</a>/<a class="move-down">Down</a>]</span>
<span class="mySpan" data-id="1" id="span-1">[<a class="move-up">Up</a>/<a class="move-down">Down</a>]</span>

JS：

var order = new Array();
$(".mySpan").each(function(){
    order.push($(this).attr('data-id'));
});

Answer 3

为了获得跨度本身：

var spanArray = []; //Contain all spans in the proper order (2,3,5,4,1)

$("span").each(function(i){
     spanArray.push($("span")[i]);
     console.log(i + ": " + this.id);
});

仅获取跨度 ID：

var spanIDsArray = []; //Contain all span ids in the proper order (2,3,5,4,1)

$("span").each(function(i){
     spanIDsArray.push(this.id);
     console.log(i + ": " + this.id);
});

Answer 4

Your variable will be look like this:

$(document).ready(function(){
 var spnArray = [];
 $("span").each(function(i)
 {
    spnArray.push($(this).attr("id"));
             });
                 console.log(spnArray);
    });

见演示：

http://jsfiddle.net/KcRTm/