mysqli中的回显脚本警报 - Echo Script Alert in mysqli

Echo Script Alert in mysqli

本文关键字：脚本 mysqli | 更新日期: 2023-11-17

嗨，我正在尝试制作一个删除脚本，当用户点击"删除"时，它运行良好，删除该帖子并弹出更改该帖子已被删除，然后重定向到他们的主页，但在将该脚本转换为mysqli后，它运行正常，但没有显示弹出更改，也没有重定向。

这是我的delete.php脚本

<?php 
$con=mysqli_connect("localhost","root","123","user");
// Check connection
if (mysqli_connect_errno())
  {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
  }
if(isset($_GET['del'])){
    $delete_id = $_GET['del'];
$delete_query = mysqli_query($con,"delete from save_data where ID='$delete_id'")
    or die(mysql_error()); 
if (mysqli_query($con, $delete_query)) {
echo "<script>alert('Image Has been Deleted')</script>";
    echo "<script>window.open('pimage.php','_self')</script>";
    }
}
?>

为什么要执行两次查询？您忘记了脚本标记中的分号

试着这样做。。

      $delete_query = mysqli_query($con,"delete from save_data where ID='$delete_id'")
or die(mysql_error()); 
  if ($delete_query) {
     echo "<script>alert('Image Has been Deleted');</script>";
     echo "<script>window.open('pimage.php','_self');</script>";
}

希望它能解决你的问题

更改：

echo "<script>alert('Image Has been Deleted')</script>";
echo "<script>window.open('pimage.php','_self')</script>";

收件人：

echo "<script>alert('Image Has been Deleted');</script>";
echo "<script>window.open('pimage.php','_self');</script>";

此外，您正在运行SQL语句两次，因此if将始终为false，因为它已被删除。因此，将$delete_query更改为：

$delete_query = "delete from save_data where ID='$delete_id'");

你能试试这个吗？你错过了脚本中的ass ;。此外，您已经多次执行mysqli_query，而不是使用一次。

    if ($delete_query) {            
           echo "<script>alert('Image Has been Deleted'); window.open('pimage.php','_self');</script>"; 
    }