Derleek Posted January 21, 2010 Share Posted January 21, 2010 Hi, I have a *very* simple jQuery/ajax script for a membership driven website. The script just checks to see if the user entered info, validates it against my database and returns 1 for a failed login attempt, and 0 if the username/password match. The function works perfectly when I run it w/o calling it from ajax. **Firefox:** everything works. **Chrome/Safari:** The query is locking everything up. When I enter valid info, there isn't even any record of a request going out (in the developer tools chrome/safari supply). When invalid info is entered the request is going out and it returns 3. I'm pretty baffled by this. Is it a "sandbox" issue? as I understand it Firefox and Chrome/safari handle XHR differently... Although don't ask me how. I am running a base install of MAMP. Could this be something on my developer environment? (I will be testing it on a live server later this afternoon) Here is my AJAX call and the php script it is calling. AJAX: $.ajax({ //I've also tried.. //url: '_modules/php/session.php', url: 'http://localhost/current/100TradeJack/httpdocs/_modules/php/session.php', type: 'POST', cache: false, timeout: 5000, data: "method=login&"+$('#loginTop').serialize(), error: function(XMLHttpRequest, ajaxOptions, thrownError) { alert("ERORR!!"); }, success: function(response){ alert(response); if(response == '0'){ alert("log user in"); window.location.replace("http://localhost/current/100TradeJack/httpdocs/trading.php"); } else if(response == '1') { alert("invalid username/password"); } else { alert("ERRROOR!"); } } }); session.php: $connection = mysql_connect("localhost","root","root") or die(mysql_error()); mysql_select_db("sandbox_members", $connection) or die(mysql_error()); if($email){ $email = mysql_real_escape_string($email); } else { return 1; } if($password){ $password = mysql_real_escape_string($password); } else { return 1; } $query = "SELECT * FROM users WHERE username='". $email . "' AND password='". $password ."'"; $result = mysql_query($query); if($row = mysql_fetch_array($result)){ return 0; } else { return 3; } //if it gets to here the username/password combo was incorrect or was not found in the db return 1; mysql_close($connection); Link to comment https://forums.phpfreaks.com/topic/189327-simple-ajax-function-causing-difficult-debug-in-chromesafari/ Share on other sites More sharing options...
Derleek Posted January 22, 2010 Author Share Posted January 22, 2010 mysql_close($conn) needed to be executed before I returned 0 or 1... duh! Link to comment https://forums.phpfreaks.com/topic/189327-simple-ajax-function-causing-difficult-debug-in-chromesafari/#findComment-1000081 Share on other sites More sharing options...
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