Question

似乎.each函数在内部.each函数中设置变量时，不会正确地循环所有项目，假设这是项目不被删除的原因。但是我用来测试循环的第二个函数表明它应该可以工作。

有什么特别的意见吗？

HTML代码：

<div id="top-menu">
    <ul id="top-menu-main-ul">
        <li class="top-menu-main-li">
            <a href="#" title="Home">Home</a>
            <div class="arrow-bg-above-ul"> </div>
            <ul class="top-menu-child-ul">
                <li class="top-li-ztyle"><a class="menu-item-2" href="index.php">Admin Home</a></li>
                <li><a class="menu-item-2" href="myaccount.php">My Account</a></li>
                <li class="bottom-li-ztyle"><a class="menu-item-0" href="logout.php">Logout</a></li>
            </ul>
        </li>

        <li class="top-menu-main-li">
            <span class="separator"><a href="#" title="Clients">Clients</a></span>
            <div class="arrow-bg-above-ul"> </div>
            <ul class="top-menu-child-ul">
                <li class="top-li-ztyle"><a class="menu-item-3" href="clients.php">View/Search Clients</a></li>
                <li><a class="menu-item-4" href="clientsadd.php">Add New Client</a></li>
                <li class="bottom-li-ztyle"><a class="menu-item-5" href="massmail.php">Mass Mail Clients</a></li>   
            </ul>
        </li>
    </ul>
</div>

jQuery代码：

var arrayOfitems = ["1","2","33","34","35","36","37","38","39","40","41","42","77"];

$(document).ready(function(){ 
    $(".top-menu-child-ul li a").each(function() {
        var thisClass = $(this).attr('class');
        var thisClassNum = thisClass.replace('menu-item-', '');
        var searchForThisNum = thisClassNum.toString();     

        if($.inArray(searchForThisNum, arrayOfitems) !== -1 )
        {
           // do nothing
        }
        else
        {
            $("." + thisClass).closest('.li').remove();
        }
    });


    // test that loop should work
    var myArray = [];

    $(".top-menu-child-ul li a").each(function() {
       myArray.push($(this).attr('class'));
    });

    alert(myArray);
});

小提琴示例： - http://jsfiddle.net/LmUPL/2/

Answer 1

您正试图找到要移除的class="li"祖先：

$("." + thisClass).closest('.li').remove();
// -------------------------^

你的意思是说：

$("." + thisClass).closest('li').remove();

找到<li>。另外，正如评论中Juan Guerrero所述，我想你想这样说：

$(this).closest('li').remove();

确保您准确定位正确的元素。

演示：http://jsfiddle.net/ambiguous/A7Ys2/2/

为什么jQuery.each不会遍历所有项目？

1 个答案: