如何检查复选框是否在 PHP 中被选中?

我正在尝试检查是否已在 PHP 中选中复选框以进行注册.但它现在不起作用.这是我目前使用的代码，用于检查是否已被检查.

if(!isset($_POST['tos']))$this->errors[] = '请接受我们的服务条款.';

这是 HTML 代码.

<标签><input type="checkbox" name="tos" value="0">我同意<a href="#">服务条款</a>和<a href="#">隐私政策</a>

完整表格代码:

 <form role="form" method="POST" action="<?php echo $_SERVER['PHP_SELF'];?>"><div class="form-group"><div class="row"><div class="col-sm-6"><label for="firstname">Firstname</label><input type="text" class="form-control" id="firstname" name="firstname" placeholder="Firstname">
<div class="col-sm-6"><label for="surname">Surname</label><input type="text" class="form-control" id="surname" name="surname" placeholder="Surname">
<div class="form-group"><label for="username">用户名</label><input type="text" class="form-control" id="ruser" name="ruser" placeholder="用户名">
<div class="form-group"><label for="email2">电子邮件地址</label><input type="email" class="form-control" id="remail" name="remail" placeholder="输入电子邮件">
<div class="form-group"><div class="row"><div class="col-sm-6"><label for="password2">密码</label><input type="password" class="form-control" id="rpass" name="rpass" placeholder="密码">
<div class="col-sm-6"><label for="password2">重复密码</label><input type="password" class="form-control" id="rpass2" name="rpass2" placeholder="密码">
<div class="checkbox"><标签><input type="checkbox" name="tos" value="0">我同意<a href="#">服务条款</a>和<a href="#">隐私政策</a>
<input type="hidden" name="secur" value="<?php echo $ip;?>"/><input type="hidden" name="token" value="<?php echo $token;?"/><button type="submit" name="register" class="btn btn-block btn-color btn-xxl">创建账户</button></表单>

这段代码有什么问题?任何帮助将不胜感激.

解决方案

if(!isset($_POST['tos']))$this->errors[] = '请接受我们的服务条款.';`

这是说如果 $_POST 中没有tos"，则抛出错误.

你真的应该用javascript检查这个客户端，在表单被发送到服务器之前.

你也可以...

和 PHP:

if(isset($_POST['tos']) && $_POST['tos']==='accepted') echo 'All good';else array_push($errors,'此处的错误代码');//添加到$error数组

更新

我不确定您是否在使用 jQuery，但只是为了更用户友好的方法:

var tos = $('#tos');var notice = $('.notice');tos.on('点击',function(){if(tos.is(':checked')){notice.text('谢谢你.');}别的{notice.text('请接受我们的 ToS 以继续.');}})

在此处查看工作示例. 如果未选中，您甚至可以一起暂停表单.顺便说一句，如果我没记错的话 checkbox 输入只发送 POST 数据 if 选中.

I am trying to check if a checkbox have been checked in PHP in order to register. But it's not working right now. This is the code I am using so far, to check if it have been checked.

if(!isset($_POST['tos']))
  $this->errors[] = 'Please accept our Terms of Service.';

And this is the HTML code.

<div class="checkbox">
<label>
<input type="checkbox" name="tos" value="0"> I agree to the <a href="#">Terms of Service</a> and <a href="#">Privacy Policy</a>
</label>
</div>

Full form code:

        <form role="form" method="POST" action="<?php echo $_SERVER['PHP_SELF'];?>">
              <div class="form-group">

              <div class="row">
                  <div class="col-sm-6">
                  <label for="firstname">Firstname</label>
                  <input type="text" class="form-control" id="firstname" name="firstname" placeholder="Firstname">
                  </div>
                  <div class="col-sm-6">
                  <label for="surname">Surname</label>
                  <input type="text" class="form-control" id="surname" name="surname" placeholder="Surname">
                  </div>
              </div>
              </div>
              <div class="form-group">
                <label for="username">Username</label>
                <input type="text" class="form-control" id="ruser" name="ruser" placeholder="Username">
              </div>
              <div class="form-group">
                <label for="email2">Email address</label>
                <input type="email" class="form-control" id="remail" name="remail" placeholder="Enter email">
              </div>
              <div class="form-group">
                <div class="row">
                  <div class="col-sm-6">
                  <label for="password2">Password</label>
                  <input type="password" class="form-control" id="rpass" name="rpass" placeholder="Password">
                  </div>
                  <div class="col-sm-6">
                  <label for="password2">Repeat password</label>
                  <input type="password" class="form-control" id="rpass2" name="rpass2" placeholder="Password">
                  </div>
                </div>
              </div>
              <div class="checkbox">
                <label>
                  <input type="checkbox" name="tos" value="0"> I agree to the <a href="#">Terms of Service</a> and <a href="#">Privacy Policy</a>
                </label>
              </div>
               <input type="hidden" name="secur" value="<?php echo $ip;?>"/>
               <input type="hidden" name="token" value="<?php echo $token;?>"/>
              <button type="submit" name="register" class="btn btn-block btn-color btn-xxl">Create an account</button>
            </form>

What's wrong with this code? Any help would be appreciated.

解决方案

if(!isset($_POST['tos']))
  $this->errors[] = 'Please accept our Terms of Service.';`

What this is doing is saying if there is no 'tos' in $_POST, throw an error.

Really you should probably check this client side with javascript, before the form is sent to the server.

You could also...

<input type="checkbox" name="tos" value="accepted" checked>

And PHP:

if(isset($_POST['tos']) && $_POST['tos']==='accepted') echo 'All good'; 
else array_push($errors,'err code here'); //to add to $error array

Update

I'm not sure if you're using jQuery or not, but just for a more user friendly approach:

var tos = $('#tos');
var notice = $('.notice');
tos.on('click',function(){
  if(tos.is(':checked')){
    notice.text('Thank you.');
  }
  else{
   notice.text('Please accept our ToS to continue.');
  }
})

See a working example here. You could even halt the form all together if it's unchecked. BTW, if I remember correctly checkbox inputs only send POST data if checked.