php 如何将php变量值传递给html表单的action属性
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how to pass php variable value to action attribute of html form
提问by MAS1
I want to pass php variable value as a action to html form. I am trying as follows, but it is not working.
我想将 php 变量值作为一个动作传递给 html 表单。我正在尝试如下，但它不起作用。
<?php
$url='test.php';
?>
<html>
<body>
<form name="upload" action="<?=$url?>" method="post" >
<input type="submit" value="submit">
</form>
</body>
</html>
All this code are in one php file.
所有这些代码都在一个 php 文件中。
回答by webbiedave
Have you tried <?php echo $url ?>If it works, then short_open_tag in the php.ini is turned off. That means you will need to either turn it on or use the long open tag <?phpthroughout your code.
有没有试过<?php echo $url ?>如果可以，说明php.ini中的short_open_tag是关闭的。这意味着您需要<?php在整个代码中打开它或使用长开放标签。
回答by rmarscher
Sounds like you need to enable short_open_tagif your example doesn't work.
如果您的示例不起作用，听起来您需要启用short_open_tag。
<?php
ini_set('short_open_tag', 'on');
$url='test.php';
?>
<html>
<body>
<form name="upload" action="<?=$url?>" method="post" >
<input type="submit" value="submit">
</form>
</body>
</html>
Alternately, write it like this:
或者，写成这样：
<?php
$url='test.php';
?>
<html>
<body>
<form name="upload" action="<?php echo $url ?>" method="post" >
<input type="submit" value="submit">
</form>
</body>
</html>
回答by nc3b
Try this
尝试这个
<form name="upload" action="<? echo $url ?>" method="post" >
回答by Jakob Kruse
Remove your single quotes:
删除您的单引号：
<form name="upload" action="<?=$url?>" method="post">
