无法获得警报弹出工作[关闭]

Question

I have this simple uploader that uploads a chosen file in a folder, I use PHP, HTML and MSSQL as my database. The pop-up script doesn't work everytime I have successfully uploaded the file. What seems to be the problem? here is a part of my code:

 //move_uploaded_file function will upload your image.  if you want to resize image before uploading see this link http://b2atutorials.blogspot.com/2013/06/how-to-upload-and-resize-image-for.html

if(move_uploaded_file($_FILES["file"]["tmp_name"],"C:\Users\Joseph\Desktop\Pics/" . $_FILES["file"]["name"]))
{
  // If file has uploaded successfully, store its name in data base
  $query_image = "insert into dbo.acc_images (image, status, acc_id) values ('".$_FILES['file']['name']."', 'display','')";

  if(sqlsrv_query($conn, $query_image))
  {
    echo '<script type="text/javascript">alert("Stored in: " . "\Users\Joseph\Desktop\Pics/" . $_FILES["file"]["name"]);</script>';
  }
  else
  {
    echo 'File name not stored in database';
  }
}

Answer 1

Your string concatenation is wrong, and it's breaking your javascript.

Try this:

echo '<script type="text/javascript">alert("Stored in: \Users\Joseph\Desktop\Pics/' . $_FILES["file"]["name"] . '");</script>';