类似这样的代码无法运行,有什么解决方法?
var tgtCol = $('td[aria-describedby=tblGrid_Subject]');
var tgtHdr = $('#tblGrid_Subject');
$(tgtHdr, tgtCol).attr('colSpan', '3');
类似这样的代码无法运行,有什么解决方法?
var tgtCol = $('td[aria-describedby=tblGrid_Subject]');
var tgtHdr = $('#tblGrid_Subject');
$(tgtHdr, tgtCol).attr('colSpan', '3');
var tgtCol = $('td[aria-describedby=tblGrid_Subject]');
var tgtHdr = $('#tblGrid_Subject');
$(tgtHdr).add(tgtCol).attr('colSpan', '3');
这个也可以工作:
var stuff = $('td[aria-describedby=tblGrid_Subject], #tblGrid_Subject');
stuff.attr('colSpan', '3');
$('td[aria-describedby=tblGrid_Subject], #tblGrid_Subject').attr('colSpan', '3');
由于前两行已经返回了一个 jQuery 对象,你可以这样写:
tgtHdr.attr('colSpan', '3');
tgtCol.attr('colSpan', '3');
你可以最初使用多重选择器:
$('td[aria-describedby=tblGrid_Subject], #tblGrid_Subject').attr('colSpan', '3');
// using merge:
$.merge(selector1, selector2) // the rest
// using each:
$([selector1, selector2, selector3, .....etc]).each(function(){
// your code here
});