进来看看吧，字符串遍历问题

Question

<span id="idList">10000,10006,10007,10008,</span>    <div id="allID">        <div class="float_left div1">            <div id="accordion_panel1" class="panel">          <h3 class="gray-top" id="10000">                    <a href="add.aspx?leibie=10000&name=新闻">新闻</a>                    <a href="add.aspx?leibie=10000&name=新闻添加">新闻添加</a>                    <a href="list.aspx?leibie=10000&name=新闻列表">新闻列表</a>                </h3>             <div>             <ol>             <li id="10006">                            <a href="add.aspx?leibie=10006&name=焦点">焦点</a>                            <a href="add.aspx?leibie=10006&name=焦点添加">焦点添加</a>                            <a href="list.aspx?leibie=10006&name=焦点列表">焦点列表</a>                        </li>             <li id="10007">                            <a href="add.aspx?leibie=10007&name=国内">国内</a>                            <a href="add.aspx?leibie=10007&name=国内添加">国内添加</a>                            <a href="list.aspx?leibie=10007&name=国内列表">国内列表</a>                        </li>             <li id="10008">                                                    <a href="add.aspx?leibie=10008&name=国际">国际</a>                            <a href="add.aspx?leibie=10008&name=国际添加">国际添加</a>                            <a href="list.aspx?leibie=10008&name=国际列表">国际列表</a>                        </li>             </ol>             </div>
根据<span id="idList">10000,10006,10007,10008,</span>里面的字符串“10000,100001...”，
遍历<div id="allID"> 里面所有的ID，如果ID在上面那个字符串里面存在，那么这个标签能够显示，否则隐藏掉例如：function reId() {
      var idlist = document.getElementById("idList").innerText;
      $("div#allId").find("[id]").each(function () {
            //这边应该如何写
      })
}     
  

Answer 1

<span id="idList">accordion_panel1,10000,10006,10007</span>    <div id="allID">        <div class="float_left div1">            <div id="accordion_panel1" class="panel">             <h3 class="gray-top" id="10000">                    <a href="add.aspx?leibie=10000&name=新闻">新闻</a>                    <a href="add.aspx?leibie=10000&name=新闻添加">新闻添加</a>                    <a href="list.aspx?leibie=10000&name=新闻列表">新闻列表</a>                </h3>                <div>                    <ol>                        <li id="10006">                            <a href="add.aspx?leibie=10006&name=焦点">焦点</a>                            <a href="add.aspx?leibie=10006&name=焦点添加">焦点添加</a>                            <a href="list.aspx?leibie=10006&name=焦点列表">焦点列表</a>                        </li>                        <li id="10007">                            <a href="add.aspx?leibie=10007&name=国内">国内</a>                            <a href="add.aspx?leibie=10007&name=国内添加">国内添加</a>                            <a href="list.aspx?leibie=10007&name=国内列表">国内列表</a>                        </li>                        <li id="10008">                                                    <a href="add.aspx?leibie=10008&name=国际">国际</a>                            <a href="add.aspx?leibie=10008&name=国际添加">国际添加</a>                            <a href="list.aspx?leibie=10008&name=国际列表">国际列表</a>                        </li>                    </ol>                </div>
$(function(){
var ids = $("#idList").text().split(","),
obj = {}
for (var i = ids.length; i--;){
obj[ids[i]] = 1;
}
$("#allID [id]").each(function(){
var $this = $(this);
if (obj[$this.attr("id")]){
$this.show();
} else {
$this.hide();
}
})
})

Answer 2

window.onload=function(){
        var idsStr=$("#idList").html().split(","),allID=$("#allID")[0];
        for(var i=0;i<idsStr.length;i++){
            if(idsStr[i])
                idsStr[i]="#"+idsStr[i];
        };
        
        $("*[id]",allID).hide();
        $(idsStr.join(","),allID).show();
    };

Answer 3

写的挺好，楼主把allID里面div那个id去掉 就能用了。