Question

例如，我有一个代码：

<table>
    <tr>
        <th>name</td>
        <th>price</td>
    </tr>
    <tr>
        <td>a</td>
        <td class="sort">5</td>
    </tr>
    <tr>
        <td>b</td>
        <td class="sort">3</td>
    </tr>
    <tr>
        <td>c</td>
        <td class="sort">8</td>
    </tr>
    <tr>
        <td>c</td>
        <td class="sort"></td>
    </tr>
    <tr>
        <td>h</td>
        <td class="sort">2</td>
    </tr>
    <tr>
        <td>p</td>
        <td class="sort">6</td>
    </tr>
    <tr>
        <td>b</td>
        <td class="sort">20</td>
    </tr>
    <tr>
        <td>b</td>
        <td class="sort"></td>
    </tr>
</table>

我希望这样排序：

<table>
    <tr>
        <th>name</td>
        <th>price</td>
    </tr>
    <tr>
        <td>h</td>
        <td class="sort">2</td>
    </tr>
    <tr>
        <td>b</td>
        <td class="sort">3</td>
    </tr>
    <tr>
        <td>a</td>
        <td class="sort">5</td>
    </tr>
    <tr>
        <td>p</td>
        <td class="sort">6</td>
    </tr>
    <tr>
        <td>c</td>
        <td class="sort">8</td>
    </tr>
    <tr>
        <td>b</td>
        <td class="sort">20</td>
    </tr>
    <tr>
        <td>c</td>
        <td class="sort"></td>
    </tr>
    <tr>
        <td>b</td>
        <td class="sort"></td>
    </tr>
</table>

我使用了这段代码：

function sortNum(a, b) {
    return 1 * $(a).find('.sort').text() < 1 * $(b).find('.price').text() ? 0 : 1;
}
function sortTheTable(){
    $(function() {
        var elems = $.makeArray($('tr:has(.price)').remove())
        elems.sort(sortNum)
        $('table#information').append($(elems));
    });
}

这可行，但问题是，输出是这样的：

<table>
    <tr>
        <th>name</td>
        <th>price</td>
    </tr>
    <tr>
        <td>c</td>
        <td class="sort"></td>
    </tr>
    <tr>
        <td>b</td>
        <td class="sort"></td>
    </tr>
    <tr>
        <td>h</td>
        <td class="sort">2</td>
    </tr>
    <tr>
        <td>b</td>
        <td class="sort">3</td>
    </tr>
    <tr>
        <td>a</td>
        <td class="sort">5</td>
    </tr>
    <tr>
        <td>p</td>
        <td class="sort">6</td>
    </tr>
    <tr>
        <td>c</td>
        <td class="sort">8</td>
    </tr>
    <tr>
        <td>b</td>
        <td class="sort">20</td>
    </tr>
</table>

空的一个登顶。我希望底部的空白。

由于

Answer 1

你有多个插件来排序它为什么要重新发明轮子。这是一个这样的插件

Link

<script type="text/javascript" src="/path/to/jquery-latest.js"></script> 
<script type="text/javascript" src="/path/to/jquery.tablesorter.js"></script> 

$("#myTable").tablesorter();

Answer 2

而不是

return 1 * $(a).find('.sort').text() < 1 * $(b).find('.sort').text() ? 1 : 0;

插入

return 1 * $(a).find('.sort').text() < 1 * $(b).find('.price').text() ? 0 : 1;

http://jsfiddle.net/E56j8/

Answer 3

看看Sorting - we're doing it wrong。用于排序内容的简单jQuery插件可用here。

关于代码的一些注意事项：

// you're binding a document ready event within a function call? 
// looks the wrong way 'round, to me
function sortTheTable(){
    $(function() {
        // 1) you probably want to use .detach() over .remove()
        // 2) "tr:has(.price)" will match ALL table rows 
        // containing an element with the class .price
        // even if they're children of different <table>s!
        // 3) $('.selector') is already "an array", at least it's sortable right away.
        // there's no need for $.makeArray() here
        var elems = $.makeArray($('tr:has(.price)').remove())
        elems.sort(sortNum)
        // "#information" is a sufficient (and more efficient) selector, 
        $('table#information').append($(elems));
    });
}

根据表数据对表行进行排序

3 个答案: