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Jquery Help
fresher_06
Member
Hi all..
I just wanted to keep all of my jquery code externally which is currently inside my main .php file.
Facing one challenge as below --
In my php file I have this jquery code --
$userid = mysql_real_escape_string($_SESSION['uid']);
var userid = <?php echo "$userid"; ?>; //getting used in the Ajax call
Now but when I keep this code in an external .js file then how do I get the userid in javascript variable . which is actually coming from $_SESSION Variable ..
Any clue ??
Comments
This is not jquery code. You do not need javascript to pass the userid. The userid is stored in a SESSION variable and can be kept on the PHP side.
@Corey .. i am using it in this way ..
$(document).ready(function() {
var username = '<?php echo "$username"; ?>';
var userid = '<?php echo "$userid"; ?>'; //getting used in the Ajax call
});
Then using the userid later as below in the ajax call
var dataString = '&cust_id='+ userid +'&cust_final_points='+ cust_final_points_val;
$.ajax({ type: "POST", url: "update_cust_results.php", data: dataString, : :- .. more code .
@fresher_06 then just do this on the javascript side....
$.ajax({ type: "POST", url: "update_cust_results.php" });
Then on update_cust_results.php hold all of the user variables in the SESSION array.
$_SESSION['username'];
$_SESSION['userid'];
$_SESSION['cust_final_points'];
$_SESSION['cust_final_points_val'];
Every time you call the PHP script those variables will still be in the SESSION array just like you left them, no need to pass them around to javascript.