Embedding jquery popup in a php function

Question

Hi I have a jquery popup script I am trying to call in a php function.

The script calling the function is:

<link rel='stylesheet' href='css/sweetalert.css'>
<script src='js/sweetalert-dev.js'></script>
<script src='https://code.jquery.com/jquery-2.1.3.min.js'></script>
<?php
include('functions/notification.php');
$notification_message = "There is a problem loading Smart Telecom at the moment.";
notify($notification_message);
?>

and the function call in notification.php:

<?php
function notify($message)
{
 $pop.= "<script>sweetAlert('". $message ."', 'Our Sincere apologies.', 'error');</script>";
 echo $pop;
}

?>

As it is, the pop up will display with an error about undefined variable $pop. If I try to disable error notifications, the popup doesn't display. if I do it without the concatenation, the popup also doesn't display.

Where am I going wrong?

UPDATE - SOLVED:

Got it to work. echoed "nbsp;", just before echoing $pop

• Epodax: That is because you are using .= in your notify function, .= means to add something to existing variable, but since the variable doesn't exist before your $pop .= line, it throws that undefined variable message. - Remove the dot from the $pop .= and you should be set.
• Noman: extending @Epodax 's comment , simply change $pop.= to $pop =
• user5898266: When i do it without the concatenation, the popup does not display
• Noman: try changing variable name.
• user5898266: changing variable name hasnt worked. it only displays when i place the concatenation, which makes no sense.

Answer 1

Ok so follow these steps to ensure the code is working, then check the pop issue:

1. Delete all data in the function to look like below:

   <?php
   function notify($message) {
    echo $message;
   } //output: data passed to $message variable should be displayed?>
   

2. if echo $message works then

   <link rel='stylesheet' href='css/sweetalert.css'>
   <!-- You must always place the jQuery first and then call functions or it will not work -->
   <script src='https://code.jquery.com/jquery-2.1.3.min.js'></script>
   <script src='js/sweetalert-dev.js'></script>
   <?php
   function notify($message) {
    $pop = "<script>sweetAlert('$message', 'Our Sincere apologies.', 'error');</script>";//refer link: http://stackoverflow.com/questions/10512452/php-using-a-variable-inside-a-double-quotes
    echo $pop;
   }?>
   

• Epodax: What have you changed and why? Code only answers rarely provide much insight for the OP and future visitors, a good answer explains what OP was doing wrong / what you have changed to make it work.
• user5898266: No luck. still not displaying popup
• Ravistm: @user5898266, Refer this for variable with different variations: w3schools.com/php/func_string_echo.asp