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选择下拉菜单项点击按钮提交跳转到另一个页面显示出数据库数据

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Release： 2016-06-23 13:35:00
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回复讨论(解决方案)

不明所以，麻烦问得时候问得具体点

楼主，你的图挂了~~~
我觉得你的意思可能是从select下拉框选择一个项，然后跳转到另一个页面，option的value作为参数？
使用jquery可以很方便的完成：$("#select").click(function(){这里写你的方法就行了})

你的意思是value作为参数，那还是要用到jQuery是吧，不用JS直接用PHP的POST是不可以吗？就是选择select的一个项后可以显示出数据库对应的字段记录打印出来到页面。

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选择下拉菜单项点击按钮提交跳转到另一个页面显示出数据库数据
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