Ajax成功函数显示jquery的双重通知如何禁用第二个通知？_JavaScript

这是我用于删除记录的代码，但是删除成功完成后。 Waqas.notification("Data Deleted Successfully","success"); 通知在页面顶部两次显示。

function DeleteUser(deleteid) {
        $('#YesDelete').click(function() {
            $.ajax({
                url: "../Ajax/CrudSlider.php",
                type: 'POST',
                data: {
                    deleteid: deleteid
                },
                success: function(data) {
                     readRecords();
                     var split = data.split("::");
                     Waqas.notification("Data Deleted Successfully","success");

                }
            });

        });

这是从数据库中删除记录的代码。

if(isset($_POST['deleteid']))
{
    $user_id = $_POST['deleteid']; 
    $selectquery1="select * from slider where slider_id = $user_id"; 
    $cm= mysqli_query($conn,$selectquery1);
    $result =mysqli_fetch_array($cm);
    $path = $result['slider_image']; 
    $deletequery = " DELETE FROM `slider` WHERE slider_id ='$user_id' ";
    unlink($path);
    if (!$result = mysqli_query($conn,$deletequery)) {    
        exit(mysqli_error());
    }
}

这是删除模型。 单击DeleteUser() 。

<div class="modal Delete-modal">
        <div class="modal-dialog">
            <div class="modal-content">
                <div class="modal-header">
                    <button type="button" class="close" data-dismiss="modal" aria-label="Close"><span aria-hidden="true">&times;</span></button>
                    <h4 class="modal-title">Warning</h4>
                </div>
                <div class="modal-body">
                    <input type="hidden" id="DeleteHiddenID">
                    <div class="form-group">
                        <label for="Identification">Are You sure to delete the record.</label>
                    </div>
                </div>
                <div class="modal-footer">
                    <button type="button" class="btn btn-success pull-right" data-dismiss="modal" id="YesDelete">Yes</button>
                    &nbsp; &nbsp;
                    <button type="button" class="btn btn-danger" data-dismiss="modal" style="margin-right: 10px;">No</button>
                </div>
            </div><!-- /.modal-content -->
        </div><!-- /.modal-dialog -->
    </div>

每次单击删除按钮时，都将单击事件绑定到YesDelete按钮。 相反，您应该只分配一次事件，因此应将其移到DeleteUser函数之外。 之后，您所需要做的就是更新发送到服务器的deleteid 。

最简单的方法是在调用DeleteUser时将ID分配给全局变量。 然后，在调用服务器时，click事件将使用该ID。

例如

var deleteid;

$(function() {

  $('#YesDelete').click(function() {

    $.ajax({
      url: "../Ajax/CrudSlider.php",
      type: 'POST',
      data: {
        deleteid: deleteid
      },
      success: function(data) {
        readRecords();
        var split = data.split("::");
        Waqas.notification("Data Deleted Successfully", "success");
      }
    });

  });

});

function DeleteUser(id) {
  deleteid = id;
}

Ajax成功函数显示jquery的双重通知如何禁用第二个通知？

问题描述

1楼