PHP 上传 - 为什么 isset($_POST['submit']) 总是 FALSE
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PHP upload - Why isset($_POST['submit']) is always FALSE
提问by a2011
I have the following code sample upload3.php:
我有以下代码示例upload3.php：
<html>
<head>
<title>PHP Form Upload</title>
</head>
<body>
<form method='post' action='upload3.php' enctype='multipart/form-data'>
Select a File:
<input type='file' name='filename' size='10' />
<input type='submit' value='Upload' />
</form>
<?php
if (isset($_POST['submit']))
{
echo "isset submit";
}
else
{
echo "NOT isset submit";
}
?>
</body>
</html>
The code always returns "NOT isset submit". Why does this happen? Because the same script upload3.php calls itself?
代码总是返回“NOTisset submit”。为什么会发生这种情况？因为同一个脚本upload3.php调用了自己？
回答by ashurexm
You do not have your submit button named:
Change
您没有将提交按钮命名为：
更改
<input type='submit' value='Upload' />
To:
到：
<input type='submit' value='Upload' name="submit"/>
回答by Michael Mac McCann
Two things:
两件事情：
You'll want to try array_key_exists instead of isset when using arrays. PHP can have some hinky behavior when using isset on an array element.
使用数组时，您需要尝试使用 array_key_exists 而不是 isset。在数组元素上使用 isset 时，PHP 可能会有一些奇怪的行为。
http://www.php.net/manual/en/function.array-key-exists.php
http://www.php.net/manual/en/function.array-key-exists.php
if (array_key_exists('submit', $_POST)) { }
if (array_key_exists('submit', $_POST)) { }
Second, you need a name attribute on your button ( "name='submit'" )
其次，您的按钮上需要一个 name 属性（ "name='submit'" ）
回答by Artefacto
Because you don't have any form element whose nameproperty is submit.
因为您没有任何name属性为submit.
Try to use var_dump($_POST)to see the keys that are defined.
尝试使用var_dump($_POST)查看定义的键。
Notice that files are an exception; they're not included in $_POST; they're stored in the filesystem and they're metadata (location, name, etc) is in the $_FILESsuperglobal.
请注意，文件是一个例外；他们不包括在$_POST；它们存储在文件系统中，并且它们的元数据（位置、名称等）位于$_FILES超全局中。
回答by CharlesLeaf
Try looking at the REQUEST_METHODand see if it's POST. It's a little bit nicer.
尝试查看REQUEST_METHOD并查看它是否是 POST。它更好一点。
回答by Kingk
<input type='submit' value='Upload' />
should be
应该
<input type='submit' value='Upload' name='subname'/>
and that subname should be in $_POST[' ']
并且该子名应该在 $_POST[' '] 中
it will look like
它看起来像
if (isset($_POST['subname']))
{
echo "isset submit";
}
else
{
echo "NOT isset submit";
}
