elexion Posted March 22, 2009 Share Posted March 22, 2009 hey everyone, i'm having some problems again with my log in scripts.i'm getting the following error message:Notice: Undefined index: gebruikersnaam in C:\wamp\www\PhP_project\controlescript.php on line 18usersname / or password is incorrect.eventhough the password and usersname is correct.<?phpsession_start ();?><!DOCTYPE html public "-//W3C//DTD XHTML 1.0 strict //EN" "http://www.w3.org/TR/xhtmll/DTD/xhtmll-strict.dtd"> <html><head><title> Roc Asa </title><link rel="stylesheet" href="stijl.css" type="text/css"></head><body><?php include ("connect.php"); $wachtwoord = ($_POST["wachtwoord"]); $query = "SELECT * from user WHERE gebruikersnaam = '{$_POST['gebruikersnaam']}' AND wachtwoord = '".$wachtwoord."'"; $result = mysql_query($query, $db) or die('the query failed'); if (mysql_num_rows($result) > 0) { echo "correct"; $rij = mysql_fetch_array($result); echo $rij['gebruikersrol']; $_SESSION['gebruikersrol']==''; header("location: index.php?content=home"); } else { echo "<p>usersname / or password is incorrect.</p>"; } ?></body></html>this is what i have so far.can anyone give me a few hints what the problem could be.thanks for the help in advance Link to comment Share on other sites More sharing options...
jeffman Posted March 22, 2009 Share Posted March 22, 2009 The script and the page that calls the script are having a disagreement. It looks like your calling page does not have an element named gebruikersnaam. Check your form or javascript (whichever one generates the POST data).You can always var_dump($_POST) to see what data really is being posted.If you still can't figure it out, show us the form or javascript that generates the POST data. Link to comment Share on other sites More sharing options...
elexion Posted March 22, 2009 Author Share Posted March 22, 2009 thanks for the help i found the problem which took me to the next problem.i want the person that visits the website to see how he's logged in his usersname.How do i make this functional so far í have this: <?phpinclude ("connect.php");?><ul><li><a href="php"></a></li><li><a href=".php"></a></li><li><a href=".php"></a></li><li><a href=".php"></a></li><li><a href=".php"></a></li><li><a href=".php"></a></li><li><a href=".php"></a></li><li><a href=".php"></a></li><li><a href=".php"></a></li><br /><?phpif($_SESSION['gebruikersrol']=='')echo "your not logged in";if($_SESSION['gebruikersrol']=='admin'){}elseecho "you are not logged in";?></ul>the error message i get is Notice: Undefined variable: _SESSION in C:\wamp\www\PhP_project\navigatie.php on line 15your not logged inNotice: Undefined variable: _SESSION in C:\wamp\www\PhP_project\navigatie.php on line 17 Link to comment Share on other sites More sharing options...
jeffman Posted March 22, 2009 Share Posted March 22, 2009 You must use session_start() before you can set or access any session data. The word 'start' is misleading. The function is not limited to the beginning of a session. Link to comment Share on other sites More sharing options...
elexion Posted March 24, 2009 Author Share Posted March 24, 2009 thank you for that advice Deirdre's Dad its working in a way.problem i have now is i'm not sure how to make the string showthe persons usersname when he logs in. i'm pretty sure it has somethingto do with a ._GET state but not sure how to use it.this is what i have right now <?phpsession_start();include ("connect.php");echo "your logged in as: ".$_SESSION['gebruikersnaam'] = '';?>so the now it says your logged in as can anyone help me out? Link to comment Share on other sites More sharing options...
sooty2006 Posted March 24, 2009 Share Posted March 24, 2009 This should work, <?phpsession_start();include ("connect.php");$username = $_SESSION['gebruikersnaam'];print "Your logged in as: $username";?> Link to comment Share on other sites More sharing options...
elexion Posted March 24, 2009 Author Share Posted March 24, 2009 if it doesnt what could be wrong? Link to comment Share on other sites More sharing options...
sooty2006 Posted March 24, 2009 Share Posted March 24, 2009 then you mean its not working,the code works as i tested before posting.you must use session_start();before you do anything else on the script,other than that you have not set the session variable so its not there to print! Link to comment Share on other sites More sharing options...
sooty2006 Posted March 24, 2009 Share Posted March 24, 2009 to set the session variable,in your first post, it shows you are not setting the session for there username! <?phpsession_start();include ("connect.php");if (user exist check) {$username = $_POST['username'];$_SESSION['gebruikersnaam'] = $username;}else{//user does not exist exit("error!");}?> Link to comment Share on other sites More sharing options...
elexion Posted March 26, 2009 Author Share Posted March 26, 2009 hello everyone i have a little problem again not with my session variables anymore though this time it's about my querys.the code is:<?php$query = "select `project`.`projectID`, `project`.`naamproject`, `vraag`.`vraagID`, `vraag`.`vraagomschrijving` from `vraag`, `project`";$vervolg = "where `project`.`projectID` = `vraag`.`projectID`";if ($_POST['naamproject'] != 'default'){ if ($vervolg == "") { $vervolg = $vervolg." where "; } else { $vervolg = $vervolg." and "; } $vervolg = $vervolg." `vraag`.`projectID` like '%".$_POST['naamproject']."%'";}$query = $query.$vervolg;echo $vraagomschrijving;?>when i run it i for some reason get this uitloggen Notice: Undefined variable: vraagomschrijving in C:\wamp\www\new_project\strings.php on line 17ingelogt als janprojectnaam: 1leerjaar: 1project ID : PhP sessieprojectnaam: 1leerjaar: 2project ID : PhP sessieverkeerde project Link to comment Share on other sites More sharing options...
justsomeguy Posted March 26, 2009 Share Posted March 26, 2009 So what's your question? The undefined variable should be pretty obvious, the other output isn't coming from that code, there aren't any echo statements to write that stuff out. Link to comment Share on other sites More sharing options...
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