将复选框状态传递给PHP [英] Passing checkbox state to PHP

问题描述

 < input type =hiddenname =check_box_1value =0/> 
< input type =checkboxname =check_box_1value =1/> 
  

这个工作正常，但是当你点击submit时，勾选复选框时，它会同时传递隐藏的值和原来的复选框值在PHP中的$ _POST变量，这可以避免吗？

我有隐藏的值，所以未勾选的复选框被传递给$ _POST变量以及勾号。

解决方案

更好的方法是删除隐藏字段，只需检查在PHP中：

  if（$ _POST ['check_box_1'] =='1'）{/ * /} 
 else {/ *做一些未挑战的事情* /} 
  

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

This works fine, however when you click on submit, and the checkbox is ticked, it passes BOTH the hidden value and the original checkbox value to the $_POST variable in php, can this be avoided?

I have the hidden value there, so that unticked checkboxes are passed to the $_POST variable as well as the ticked ones.

解决方案

The better approach is to remove the hidden field, and simply have a check in PHP:

if ($_POST['check_box_1']=='1') { /*Do something for ticked*/ }
else { /*Do something for unticked*/ }

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