9tglhe6nwj Posted February 23, 2011 Share Posted February 23, 2011 I believe that my problem is with the isset on line 7 and 39, but I do not know of anyway around it. Can someone enlighten me. <?php session_start(); if (isset($_GET['logout'])){ unset($_SESSION['logged_in']); } ?> <html> <head> <title>Add a New Car!</title> <?php if (isset($_POST['submit'])) { //print_r($_POST); $password = md5($_POST['password']); //echo'<p>'.$password.'</p>'; $username = ($_POST['username']); //echo'<p>'.$username.'</p>'; } $db = new PDO('sqlite:login.sqlite'); $login = 'SELECT * FROM login WHERE userid="'.$username.'" AND password="'.$password.'";'; //echo'<p>'.$login.'</p>'; $attempt = $db->query($login); foreach ($attempt as $row) { //print_r($row); $_SESSION['logged_in'] = $username; } ?> </head> <body> <?php if (!isset($_SESSION['logged_in'])){ ?> <form method="post" action="Db_login.php"> <p>Username:<br /><input type="text" name="username" /> </p> <p>Password:<br /><input type="password" name="password" /></p> <p><input type="submit" name="submit" value="Log In!" /></p> </form> <?php } else { ?> <?php if(isset($_POST['submit'])) { print_r($_POST); $make = ($_POST['make']); $model = ($_POST['model']); $year = ($_POST['year']); $fuel = ($_POST['fuel_type']); $vin = ($_POST['vin_number']); $reg = ($_POST['reg_number']); } $db = new PDO('sqlite:cars.sqlite'); $sqlCars = 'INSERT INTO cars ( ) VALUES("'.$make.'","'.$model.'","'.$year.'","'.$fuel.'","'.$vin.'","'.$reg.'");'; echo'<p>'.$sqlCars.'</p.>'; ?> <form method="post" action="Db_new_Cars.php" enctype="multipart/form-data"> <p>Make:<br /><input type="text" name="make" /> </p> <p>Model:<br /><input type="text" name="model" /></p> <p>Year:<br /><input type="text" name="year" /></p> <p>Fuel Type:<br /><input type="text" name="fuel_type" /></p> <p>Vin Number:<br /><input type="text" name="vin_number" /></p> <p>Reg Number:<br /><input type="text" name="reg_number" /></p> <p>Choose a image to upload:<br /> <input type="file" name="myfile"></input></p> <p><input type="submit" name="submit" value="Submit!" /></p> </form> <h1> Protected Content!</h1> <p>You are logged in as <em><?php echo $_SESSION['logged_in']; ?> </em></p> <p><a href="Db_login.php?logout=true">Log out!</a></p> <p><a href="New_Cars.php">Add a new car</a></p> <p><a href="Book_Out_Cars.php">Book out a car</a></p> <?php } ?> </body> </html> Link to comment https://forums.phpfreaks.com/topic/228640-isset/ Share on other sites More sharing options...
Pikachu2000 Posted February 23, 2011 Share Posted February 23, 2011 Neither do we, unless you explain what the problem actually is. Link to comment https://forums.phpfreaks.com/topic/228640-isset/#findComment-1178861 Share on other sites More sharing options...
9tglhe6nwj Posted February 23, 2011 Author Share Posted February 23, 2011 Ohhh sorry I didn't think to explain >.< /facepalm. Anyway the problem is I login, then when I go to submit the second form it refreshes the page and I'm logged out again. Link to comment https://forums.phpfreaks.com/topic/228640-isset/#findComment-1178862 Share on other sites More sharing options...
btherl Posted February 23, 2011 Share Posted February 23, 2011 You've got two buttons called "submit" there. Rather than using isset(), you might want to check if $_POST['submit'] == 'Log In!' or if it == 'Submit!'. Also those forms there point to two different scripts, though it looks like this script handles both of them. What's going on there? Link to comment https://forums.phpfreaks.com/topic/228640-isset/#findComment-1178882 Share on other sites More sharing options...
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