为什么复选框状态并不总是传递给 PHP 脚本?

问题描述

我有一个 HTML 表单:

I have an HTML form:

<form action='process.php' method='post'>
  <input type='checkbox' name='check_box_1' /> Check me!<br>
</form>

这是 PHP 脚本 process.php 中的一段:

Here is a section from the PHP script process.php:

echo (isset($_POST['check_box_1']))?'Set':'Not set';

设置复选框时脚本的输出为

The output of the script when the checkbox is set is

设置

但是当没有设置复选框时，输出是:

But when the checkbox is not set, the output is:

未设置

这是为什么?这似乎是一个非常糟糕的设计，因为我的 PHP 脚本检查了许多 $_POST 变量以确保它们被传递给脚本.当 $_POST['check_box_1'] 值未设置时，我怎么知道是脚本未能传递值还是复选框未设置?

Why is this? This seems like a very poor design because my PHP script checks a number of $_POST variables to make sure they were passed along to the script. When the $_POST['check_box_1'] value is not set, then how do I know whether the script failed to pass along the value or the checkbox was just not set?

推荐答案

如果你想克服这个设计特点，试试这个:

If you want to overcome this design feature, try this:

<input type="hidden" name="check_box_1" value="0" />
<input type="checkbox" name="check_box_1" value="1" />

这样，您将始终在回调页面中设置 $_POST['check_box_1'] ，然后您可以检查其值以查看他们是否检查过它.不过，这两个输入必须按该顺序排列，因为后者优先.

This way, you'll always have $_POST['check_box_1'] set in the callback page, and then you can just check its value to see whether they checked it or not. The two inputs have to be in that order though, since the later one will take precedence.

这篇关于为什么复选框状态并不总是传递给 PHP 脚本?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持跟版网！