警报在PHP代码中不起作用

Question

Why my alert box is not working. Anyone can help me for find out my mistake?

I tried various way. But its not working. finally I try this way. Even this is also not working.

My aim is, if multi query executed successfully means, alert the success message then reload to the another page.

Please Help me.

<?php
if (!isset($_SESSION)) {
  session_start();
}

if (isset($_GET['insertQuery'])) {
  include("connection/connectionMysqli.php");

  if (!mysqli_multi_query($conn,$_SESSION['insertQuery'])) {
    echo "<script>alert('Faild Due to server Problem.!!!');window.refresh(true);</script>";
  } else {
    unset($_SESSION['insertQuery']);
    echo "<script>alert('Success');</script>";
    header("Location:bluk_resource_booking.php?successInsert=1");
    exit;
  }
}
?>

Answer 1

Using header("Location: ...") will cause PHP to respond with a 302 Found status, rather than a 200 Success status.

A 3xx status will cause the browser to do a redirect, ignoring anything you send in the body (including JavaScript).

---

Instead you can alert and then redirect using JavaScript.

} else {
  unset($_SESSION['insertQuery']);
  echo "<script>alert('Success'); window.location.href = 'bluk_resource_booking.php?successInsert=1';</script>";
  exit;
}