使用php删除;来自数据库的ajax，js

Question

用户登录，他想删除他的状态。

<p><?php echo $status; ?></p> //this shows status
<a href="" id="<?php echo $upid; ?>" class="delete">Delete</a> 

这里是javascript ajax代码

<script type="text/javascript" >
$(function() {

$(".delete").click(function(){
var del_id = element.attr("id");
var info = 'id=' + del_id;
if(confirm("Sure you want to delete this update? There is NO undo!"))
{
$.ajax({
type: "POST",
url: "delete.php",
data: info,
success: function(){
}
});
$(this).parents(".record").animate({ backgroundColor: "#fbc7c7" }, "fast")
.animate({ opacity: "hide" }, "slow");
}
return false; 
});
});
</script>

这是我的delete.php

<?php 
include("includes/db_connect.php"); 

if(isset($_GET['id'])){
$delete_id = $_GET['id'];

$delete_query = "delete from user_posts where upid='$delete_id'";

mysqli_query($con, $delete_query);
}

?>

用户个人资料链接是profile.php？id = 1此ID是用户ID ..不是帖子ID ..当我将鼠标移动到删除按钮时，在底部显示个人资料ID，而不是删除ID。帮助我。

http://prntscr.com/3n517j

Answer 1

type: "POST",

将其更改为“GET”。

否则你将无法在$ _GET中找到该值。

Answer 2

更改为，

if(isset($_POST['id'])){
$delete_id = $_POST['id'];