Question

单击“确认”按钮后，我想执行这些操作。但我的代码不会在`if（isset（$ _ POST ['confirm']））{..}之后运行我真的会帮助你。

echo "<div class='container'>
    <form action='confirm.php' method='post'>
    <center><button type='button' name='confirm' value='confirm' class='btn btn-info btn-lg' data-toggle='modal' data-target='#myModal'>Confirm</button></center>
    </form>
    <div class='modal fade' id='myModal' role='dialog'>
    <div class='modal-dialog'>";


 if (isset($_POST['confirm'])){

        $query2 = "SELECT status FROM reserver WHERE picker='$picking' AND droper='$droping'";
        $result1 = mysqli_query($conn, $query2);
        $row1=mysqli_fetch_assoc($result1);
        $check_status=$row1['status'];
        if($check_status=='available')
        {

         $query3 = "UPDATE reserver SET status='reserved' WHERE picker='$picking' AND droper='$droping'";
         if(mysqli_query($conn, $query3)){



 echo " <div class='modal-content'>
    <div class='modal-header'>
      <button type='button' class='close' data-dismiss='modal'>&times;</button>
      <h4 class='modal-title'>Confirmation</h4>
    </div>
    <div class='modal-body'>
      <p>Confirmed Sucessfully!</p>
    </div>
    <div class='modal-footer'>
      <button type='button' class='btn btn-default' data-dismiss='modal'>Close</button>
    </div>
  </div>";
         }
        }
      else  {
            echo " <div class='modal-content'>
    <div class='modal-header'>
      <button type='button' class='close' data-dismiss='modal'>&times;</button>
      <h4 class='modal-title'>Notice</h4>
    </div>
    <div class='modal-body'>
      <p>Unavailable Vehicle!</p>
    </div>
    <div class='modal-footer'>
      <button type='button' class='btn btn-default' data-dismiss='modal'>Close</button>
    </div>
  </div>";
        }

 }

编辑：现在我有另外一个用于显示不可用车辆的案例

Answer 1

您的解决方案无法打开带有成功消息的模式。您可以尝试这样的Pastebin Link

<?php
$status = false;
if (isset($_POST['confirm'])) {

    $check_status = 'available';
    $query2 = "SELECT status FROM reserver WHERE picker='$picking' AND droper='$droping'";
    $result1 = mysqli_query($conn, $query2);
    $row1=mysqli_fetch_assoc($result1);
    $check_status=$row1['status'];
    if($check_status=='available')
    {
        $query3 = "UPDATE reserver SET status='reserved' WHERE picker='$picking' AND droper='$droping'";
        if(mysqli_query($conn, $query3)) {
            $status = true;
        }
    }
}
?>
<div class='container'>
    <form action='confirm.php' method='post'>
        <center><button type='submit' name='confirm' value='confirm'>Confirm</button></center>
        <button type='button' name='confirm' value='confirm' class='btn btn-info btn-lg' data-toggle='modal' data-target='#myModal' style="display:none;">Confirm</button>
    </form>
    <div class='modal fade' id='myModal' role='dialog'>
        <div class='modal-dialog'>
            <div class='modal-content'>
                <div class='modal-header'>
                    <button type='button' class='close' data-dismiss='modal'>&times;</button>
                    <h4 class='modal-title'>Confirmation</h4>
                </div>
                <div class='modal-body'>
                    <p>Confirmed Sucessfully!</p>
                </div>
                <div class='modal-footer'>
                    <button type='button' class='btn btn-default' data-dismiss='modal'>Close</button>
                </div>
            </div>
        </div>
    </div>
</div>
<?php if( $status == true ): ?>
    <script>
        $('.btn.btn-info').trigger('click');
    </script>
<?php endif; ?>

Answer 2

但是如果你想提交数据并通过php处理它，你想要做什么，你需要使用<input type='submit'/>而不是按钮。

<input type="button" />使用javascript并且不会提交表单。 <input type="submit" />将提交表单。

但是AJAX就是为了这个，你只需要学习一些基础知识就可以了jQuery

所以这是示例如何工作

您首先需要创建一个简单的HTML表单

如果你想附加一些数据

<form method="POST" id="my_form">
<input type="hidden" name='test'/>
<input type="button" id="submit"/>
</form>

或者只是创建一个简单的按钮

<button id="click me" id="click_me"></button>

然后创建另一个HTML元素以显示用户响应状态

<div id="response"></div>

不可。所以现在我们完成了我们的HTML代码，让我们去编写一些jQuery代码

$(document).ready(function() {
$("#click_me").click(function() {
$.ajax({
url : "your_php_file.php", // file path you want to send data
type : "POST",
data : {},
success : function(data) {
     $("#response").html("Confirmed Sucessfully!");
}
});
});
});

最后一件事是你的PHP代码

<?php
if() {
// your php code
}else{
header("HTTP/2.0 400 Bad request");
}
?>

这只是一个简单的解释，如何处理AJAX请求。您需要了解有关此内容的更多信息才能构建安全且有用的Web应用程序

在模态中单击按钮后如何执行代码

2 个答案: