van*_*van 8 jquery jquery-selectors
我有以下代码:
var aggrHTML = $('TBODY#aggr > tr > td > table > tbody > tr > td > nobr > b');
var aggrText = aggrHTML.text();
var newText = "Total" + aggrText.substring(3);
aggrHTML.html(newText);
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我想做的是改变上面所以我选择id为'aggr'的所有元素.
非常感谢,导航
Sco*_*den 13
var aggrHTML = $('TBODY[id*=aggr] > tr > td > table > tbody > tr > td > nobr > b');
var aggrText = aggrHTML.text();
var newText = "Total" + aggrText.substring(3);
aggrHTML.html(newText);
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实现这一目标的最简单方法是
var aggrHTML = $("[id*=aggr]");
var aggrText = aggrHTML.text();
var newText = "Total" + aggrText.substring(3);
aggrHTML.html(newText);
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如果由"所有元素"表示当然是所有DOM元素.