http://jsfiddle.net/9sKwJ/66/排序行對與的tablesorter
tr.spacer { height: 40px; }
$.tablesorter.addWidget({
id: 'spacer',
format: function(table) {
var c = table.config,
$t = $(table),
$r = $t.find('tbody').find('tr'),
i, l, last, col, rows, spacers = [];
if (c.sortList && c.sortList[0]) {
$t.find('tr.spacer').removeClass('spacer');
col = c.sortList[0][0]; // first sorted column
rows = table.config.cache.normalized;
last = rows[0][col]; // text from first row
l = rows.length;
for (i=0; i < l; i++) {
// if text from row doesn't match last row,
// save it to add a spacer
if (rows[i][col] !== last) {
spacers.push(i-1);
last = rows[i][col];
}
}
// add spacer class to the appropriate rows
for (i=0; i<spacers.length; i++){
$r.eq(spacers[i]).addClass('spacer');
}
}
}
});
$('table').tablesorter({
widgets : ['spacer']
});
<table id="test">
<thead>
<tr>
<th>Name</th>
<th>Number</th>
<th>Another Example</th>
</tr>
</thead>
<tbody>
<tr>
<td>Test4</td>
<td>4</td>
<td>Hello4</td>
</tr>
<tr>
<td colspan="3">Test4</td>
</tr>
<tr>
<td>Test3</td>
<td>3</td>
<td>Hello3</td>
</tr>
<tr>
<td colspan="3">Test3</td>
</tr>
<tr>
<td>Test2</td>
<td>2</td>
<td>Hello2</td>
</tr>
<tr>
<td colspan="3">Test2</td>
</tr>
<tr>
<td>Test1</td>
<td>1</td>
<td>Hello1</td>
</tr>
<tr>
<td colspan="3">Test1</td>
</tr>
</tbody>
</table>
此排序只是我想,如果你在第一列進行排序的方式,但其他兩列別我不會保持同樣的配對'tr'排序即時尋找。
對此有何幫助?
我更新了小提琴鏈接incase你去了第一個,再次檢查。 – user1659043