php - 如果已选中，如何将复选框值插入数据库表？

Question

我的 HTML 表单上有这两个复选框。

<div class="label-input-wrapper pickup">    
        <div class="form-label">I need Airport pick-up</div>
        <div class="form-input">
            <input type="checkbox" name="pick_up_yes" value="yes" />Yes 
            <input type="checkbox" name="pick_up_no" value="no" />No
            <div class="error-msg">
                <?php if(isset($errors['pick_up_no'])) { echo '<span style="color: red">'.$errors['pick_up_no'].'</span>'; } ?>
            </div>
        </div>
</div>  

在表单提交时，值是这样保存的:

$pick_up_yes = $_POST["pick_up_yes"];
$pick_up_no = $_POST["pick_up_no"];

和验证:

if (($pick_up_yes != yes) && ($pick_up_no != no)){
            //Blank string, add error to $errors array.        
            $errors['pick_up_no'] = "Please let us know your airport pick up requirement!";
        }

我想在提交表单时将此Yes 或No 保存到我的数据库中。 air_port_pick_up 的我的数据库列名称。 如何将其保存到数据库中？如果选中 pick_up_yes 复选框，那么我想将 Yes 保存到 air_port_pick_up 如果选中 pick_up_no 复选框，那么我想将 No 保存到 air_port_pick_up

Answer 1

您必须为两个输入提供相同的名称属性 (name="pickup")

<p><input type="checkbox" name="pick_up" value="yes" />Yes</p>
<p><input type="checkbox" name="pick_up" value="no" />No</p>

像下面这样添加选中的属性

<p><input type="checkbox" name="pick_up" value="yes" checked />Yes</p>
<p><input type="checkbox" name="pick_up" value="no" />No</p>

跳过这一步

   if (($pick_up_yes != yes) && ($pick_up_no != no)){
        //Blank string, add error to $errors array.        
        $errors['pick_up_no'] = "Please let us know your airport pick up requirement!";
    }

现在您可以存储值(value):

$value = $_POST['pickup'];