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php如何通过jquery获取 $_FILES['file']['name']的值?

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Jun 23, 2016 pm 02:04 PM

在客户端放一个对象，如何用jquery 的post方法传值并在服务器端php文件里获取对象 $_FILES['file']['name']的值

回复讨论(解决方案)

用jquery forms 中的 ajaxSubmit

直接form提交就能获得：

用jquery forms 中的 ajaxSubmit

试过ajaxSubmit，一提交就提交好多次不知道啥回事，比如上传一个附件，结果一提交就提交了好几个同样的副本

直接form提交就能获得：

需要用jquery，需要从A页面向B页面无刷新提交

我还在研究 php中， jq 还需要打好基础先

上传文件关jQuey什么事。

估计楼主是想在上传图片的时候，让用户看到图片，通过JS是无法获取上传控件里面本地文件的地址的，你可以将上传部分跟表单提交部分分离，将上传放到一个iframe里面，然后先提交iframe,并用cookie保存图片的的路劲，再提交表单然后插入数据库。

同楼主一样的问题纠结中，php中$_FILES 取不到js文件post过来的值貌似只能由php直接post过来才能取到

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