存在隐藏的输入类型时，不返回任何POST数据 [英] No POST data being returned when hidden input type is present

问题描述

我认为我的代码中有错误，或者我的PHP或Apache设置不正确.

I think that there is either an error in my code, or my PHP or Apache is set up incorrectly.

提交表单中包含隐藏字段的表单时，我的$ _POST数组中没有任何数据...

When I submit a form with a hidden field in it, I do not get any data in my $_POST array...

当我注释掉代码中的隐藏字段时，POST数据将正确返回...

When I comment out the hidden field in my code, the POST data is returned correctly...

HTML表单

<form action='/utils/login.php ' method='POST'>
<table>
    <tr>
        <td colspan='2'>
            Login
        </td>
    </tr>
    <tr>
        <td>
            Username
        </td>
        <td>
            <input type='text' name='userid' value='' size='12' />
        </td>
    </tr>
    <tr>
        <td>
            Password
        </td>
        <td>
            <input type='password' name='password' size='12' />
        </td>
    </tr>
    <tr>
        <td>
            <input type='hidden' name='formtype' value='login' />
        </td>
    </tr>
    <tr>
        <td>
            <input type='submit' value='Submit' />
        </td>
    </tr>
</table></form>

这是在PHP中处理它的代码...

Here is the code that is processing it in PHP...

foreach ($_POST as $var => $value) {
     echo "$var = $value<br>";
} 

我在服务器上使用PHP 5和Apache 2.2.

I am using PHP 5 and Apache 2.2 on my server.

有什么想法吗?

编辑...

我将范围缩小到了这个...

I have narrowed it down to this...

$command = $_POST['formtype'];

当我从$ _POST删除@符号时，出现以下错误...

When I removed the @ sign from my $_POST, I am getting the following error...

注意:未定义的变量:第17行的C:\ webroot \ utils \ login.php中的表单类型

Notice: Undefined variable: formtype in C:\webroot\utils\login.php on line 17

如果我将该行注释掉，则POST数据将毫无问题地传递到程序中.

If I comment out that line, the POST data is passed into the program without a problem.

推荐答案

您是否尝试过将隐藏的输入从表中取出并放在开始表单标记之后?

Have you tried taking the hidden input out of the table and placing it right after the opening form tag?

您还可以使用:

var_dump($_POST);

...查看帖子变量.

...to view the post variables.

此外，如果正在动态创建任何输入或POST变量中可能缺少这些输入，则可以使用:

Also, if any inputs are being dynamically created or might be missing from the POST variables... you can use:

variable = 'default';
if(isset($_Post['variable'])) $variable = $_POST['variable'];

...动态设置可能存在或不存在的变量.

...to dynamically set variables that could be there or not.

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