如何在删除时刷新通知计数并删除 parent

Question

当通过 ajax 调用删除时，如何从视图中删除该内容？

这是我的 ajax 电话，当它删除时我正在从那里删除作业。

parent 必须删除并且
作业通知计数必须刷新（我的作业 (3) 它应该更改为 2，因为当我注销并登录时，数据库中现在只有 2 个作业，它是 2。）

这是我的 ajax 控制器代码：

elseif(isset($_POST['res'])&& ($_POST['res'] =='no'))
{
    $jobId=$_POST['jobId'];
    $updateStatus=$conn->query("UPDATE r_job_invitations SET inv_res='2' WHERE id_job='".$jobId."' ");
    if ($updateStatus) {
        $response['message'] = "<strong>Success!</strong> Job Rejected.";
        $response['success'] = true;
        } else {
        $response['message'] = "<strong>Warning!</strong> There is an error in Job Rejection please try again";
        $response['success'] = false;
    }
    echo json_encode($response);
    exit;
}

我的ajax呼叫和发送请求：

<a href="javascript:;" class="btn btn-default" onclick="jobResponse('no',<?php echo $myjobs['id_job'];?>)">Reject And Delete</a>

<script type="text/javascript">
    function jobResponse(res,jobId){
        var frm_data = { res : res,
            jobId : jobId
        }
        $.ajax({
            method: "POST",
            url: 'inv-controller.php',
            data: frm_data,
            dataType: "json",
            success: function (response) {
                if (response["success"] == true)
                {   
                    $("#success-message").show();
                    $("#success-message").html(response["message"]);
                    } else {
                    $("#warning-message").show();
                    $("#warning-message").html(response["message"]);
                }
            },
            error: function (request, status, error) {
                $("#warning-message").show();
                $("#warning-message").html("OOPS! Something Went Wrong Please Try After Sometime!");
            }
        });
    }
</script>

最后这个作业计数会话来自登录控制器

登录时，我正在检查作业在会话中的计数，如下所示：

$Query = $conn->query("SELECT count(*) as notificationCount FROM r_job_invitations where email='".$_SESSION['user_email']."' and inv_res=0") or die(mysqli_error());
$notification = mysqli_fetch_assoc($Query);     
$_SESSION['notificationCount'] = $notification['notificationCount'];
echo BASE_URL."my-profile.php";

我想更新此会话计数。我该怎么做呢？

Answer 1

在您的 ajax 控制器页面上：

if ($updateStatus) {
    $response['message'] = "<strong>Success!</strong> Job Rejected.";
    $response['success'] = true;
    $response['newCount'] = $_SESSION['notificationCount']+1;
    $_SESSION['notificationCount']++;
......

然后在你的 jquery

if (response["success"] == true)
            {  
                $("#success-message").show();
                $("#success-message").html(response["message"]);
                $('selector_that_has_your_old_count').html(response["newCount"]);

Answer 2

为了删除父 class 这些是我所做的更改：查看和回复：

<div class="row" id="jobrow<?php echo $myjobs['id_job'];?>">
$("#jobrow"+jobId).remove();

更新计数：查看页面：

<span class="badge" id="notificationcount">

回复：如果（响应["success"] == 真） {
$("#success-message").show(); $("#success-message").html(响应["message"]); $("#notificationcount").html(响应["notificationCount"]); }

控制器页面：

$Query = $conn->query("SELECT count(*) as notificationCount FROM r_job_invitations where email='".$_SESSION['user_email']."' and inv_res=0") or die(mysqli_error());
$notification = mysqli_fetch_assoc($Query);     
$_SESSION['notificationCount'] = $notification['notificationCount'];            

if ($updateStatus) {
    $response['message'] = "<strong>Success!</strong> Job Rejected.";
    $response['notificationCount'] = $_SESSION['notificationCount'];            
    $response['success'] = true;
    } else {
    $response['message'] = "<strong>Warning!</strong> There is an error in Job Rejection please try again";
    $response['success'] = false;
}
echo json_encode($response);
exit;
}

如何在删除时刷新通知计数并删除 parent

How can i refresh notification count and remove parent when it is deleted

php

jquery

ajax

parent-child

removechild