Javascript 如何在php上获取html输入值

Javascript 如何在php上获取html输入值,javascript,php,html,Javascript,Php,Html,我需要存储一些可以从javascript中提取的变量,稍后在提交表单时需要在php上使用这些变量 <form id=form_id enctype="multipart/form-data" encoding='multipart/form-data' method='post' > <input class=" button1" type="submit" name="Submit" value="Upload"> <input type="hidden" n

我需要存储一些可以从javascript中提取的变量,稍后在提交表单时需要在php上使用这些变量

<form id=form_id enctype="multipart/form-data" encoding='multipart/form-data' method='post' >
 <input class=" button1" type="submit" name="Submit" value="Upload">
 <input type="hidden" name="videoname_var" id="videoname_var" />
</form>
所以我在html上创建了输入字段,比如

<form id=form_id enctype="multipart/form-data" encoding='multipart/form-data' method='post' >
 <input class=" button1" type="submit" name="Submit" value="Upload">
 <input type="hidden" name="videoname_var" id="videoname_var" />
</form>
HTML

  <input type="hidden" name="videoname_var" id="videoname_var" method="get"/>
<form id=form_id enctype="multipart/form-data" encoding='multipart/form-data' method='post' >
 <input class=" button1" type="submit" name="Submit" value="Upload">
 <input type="hidden" name="videoname_var" id="videoname_var" />
</form>
稍后提交时,我将从php访问如下值:

$videoname = $_GET['videoname_var'];
<form id=form_id enctype="multipart/form-data" encoding='multipart/form-data' method='post' >
 <input class=" button1" type="submit" name="Submit" value="Upload">
 <input type="hidden" name="videoname_var" id="videoname_var" />
</form>
但是$videoname的值总是空的

<form id=form_id enctype="multipart/form-data" encoding='multipart/form-data' method='post' >
 <input class=" button1" type="submit" name="Submit" value="Upload">
 <input type="hidden" name="videoname_var" id="videoname_var" />
</form>
这个方法有什么问题吗

<form id=form_id enctype="multipart/form-data" encoding='multipart/form-data' method='post' >
 <input class=" button1" type="submit" name="Submit" value="Upload">
 <input type="hidden" name="videoname_var" id="videoname_var" />
</form>
编辑:

<form id=form_id enctype="multipart/form-data" encoding='multipart/form-data' method='post' >
 <input class=" button1" type="submit" name="Submit" value="Upload">
 <input type="hidden" name="videoname_var" id="videoname_var" />
</form>
这是来自

<form id=form_id enctype="multipart/form-data" encoding='multipart/form-data' method='post' >
 <input class=" button1" type="submit" name="Submit" value="Upload">
 <input type="hidden" name="videoname_var" id="videoname_var" />
</form>
并在“提交”按钮上单击

<form id=form_id enctype="multipart/form-data" encoding='multipart/form-data' method='post' >
 <input class=" button1" type="submit" name="Submit" value="Upload">
 <input type="hidden" name="videoname_var" id="videoname_var" />
</form>
这个php正在执行

<form id=form_id enctype="multipart/form-data" encoding='multipart/form-data' method='post' >
 <input class=" button1" type="submit" name="Submit" value="Upload">
 <input type="hidden" name="videoname_var" id="videoname_var" />
</form>
if (isset($_FILES['file']))
{
 $videoname = $_GET['videoname_var'];
}

此表单不可输入

<form id=form_id enctype="multipart/form-data" encoding='multipart/form-data' method='post' >
 <input class=" button1" type="submit" name="Submit" value="Upload">
 <input type="hidden" name="videoname_var" id="videoname_var" />
</form>
method="get"
像这样

<form id=form_id enctype="multipart/form-data" encoding='multipart/form-data' method='post' >
 <input class=" button1" type="submit" name="Submit" value="Upload">
 <input type="hidden" name="videoname_var" id="videoname_var" />
</form>
<form method="get" action="">
  <input type="hidden" value="videoname_var value" name="videoname_var" id="videoname_var" />
</form>

您必须在
表单
元素
中设置表单方法,但不能在任何其他表单的元素中设置

<form id=form_id enctype="multipart/form-data" encoding='multipart/form-data' method='post' >
 <input class=" button1" type="submit" name="Submit" value="Upload">
 <input type="hidden" name="videoname_var" id="videoname_var" />
</form>
表单
元素的
方法
属性告诉浏览器如何将表单数据发送到web服务器:

<form id=form_id enctype="multipart/form-data" encoding='multipart/form-data' method='post' >
 <input class=" button1" type="submit" name="Submit" value="Upload">
 <input type="hidden" name="videoname_var" id="videoname_var" />
</form>
<form method="get">
...
</form>

...

请注意,
get
是默认值,并以名称/值对的形式将表单数据附加到表单的动作URL中:URL?name=value&name=value

这是错误的
如果从JS设置值,则必须以某种方式将其发送到PHP。你在做表格提交吗,或者AJAX?不,我没有提交。你真的在javascript中发出GET请求吗?发布一个更完整的代码示例:如果你使用javascript,那么表单和提交函数。如果你使用GET,你不必设置方法,因为GET是表单的默认方法。我不确定为什么会被否决;考虑到另一个得到的答案是相同的。顺便说一句,这里不是我的dv。编辑:下一票似乎已被撤回。如果使用GET,则不必设置方法,因为GET是表单的默认方法。
<form id=form_id enctype="multipart/form-data" encoding='multipart/form-data' method='post' >
 <input class=" button1" type="submit" name="Submit" value="Upload">
 <input type="hidden" name="videoname_var" id="videoname_var" />
</form>