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jquery/js/php的问题?

WBOY
Release: 2016-06-20 12:43:51
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点击一个button弹出一个新窗体;
判断这个窗体是否存在,如果存在:再点击这个button不会再弹出这个窗体;如果不存在:直接弹出;
求一段代码……


回复讨论(解决方案)

如果你的窗体是 div 模拟的,那么只是 display=block
无需判断,设置多少次都一样

如果你的窗体是 open 打开的弹窗,那么只要指定窗口名(open 的第二个参数)
无需判断,总是打开在同一窗口

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