shatteredcode Posted May 26, 2008 Share Posted May 26, 2008 I'm trying to set up a login script with the users coming form a database. I can't, however, seem to extract that information. What's wrong with my script? <?php INCLUDE "connect.php"; $host = "host"; $database = "database"; $user = "user"; // I changed these for the post, they're right in my script $pass = "pass"; ConnectToDatabase($database, $user, $pass, $host); /**********/ //Select our table. $query = "SELECT user, pass FROM users"; //Query the table (Get the information from it.) $result = mysql_query($query); //Extracts the array from the table $row = mysql_fetch_array($result); $database_user == $row['user']; echo ($database_user); print "hello, why don't you login? <br />"; ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Login</title> </head> <body> <form id="login" name="login" method="post" action="login.php"> <label> User <input type="text" name="user" id="user" /> </label> <p> Password <input type="password" name="pass" id = "pass" /> </p> <input type="submit" id="submit" name="sumbit" /> </form> </body> </html> <?php if ($typed_user == $database_user) { echo('Worked!'); } else { echo(' Didn\'t Work!'); } ?> I'm not sure of my error. BTW: I'm only checking the user right now, not the password. This is just a test. Thanks. Link to comment https://forums.phpfreaks.com/topic/107265-solved-quick-database-question/ Share on other sites More sharing options...
jscix Posted May 26, 2008 Share Posted May 26, 2008 <?php INCLUDE "connect.php"; $host = "host"; $database = "database"; $user = "user"; // I changed these for the post, they're right in my script $pass = "pass"; ConnectToDatabase($database, $user, $pass, $host); $query = "SELECT user, pass FROM users WHERE `user` = '$typed_user'"; $result = mysql_query($query); $row = mysql_fetch_array($result); $database_user == $row['user']; echo($database_user); print "hello, why don't you login? <br />"; ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Login</title> </head> <body> <form id="login" name="login" method="post" action="login.php"> <label> User <input type="text" name="user" id="user" /> </label> <p> Password <input type="password" name="pass" id = "pass" /> </p> <input type="submit" id="submit" name="sumbit" /> </form> </body> </html> <?php if ($typed_user == $database_user) { echo('Worked!'); } else { echo(' Didn\'t Work!'); } ?> Link to comment https://forums.phpfreaks.com/topic/107265-solved-quick-database-question/#findComment-550008 Share on other sites More sharing options...
Barand Posted May 26, 2008 Share Posted May 26, 2008 $query = "SELECT user, pass FROM users WHERE `user` = '$typed_user'"; $typed_user has not been defined. You need to get it from the posted data. $database_user == $row['user']; to assign a value, use "=" and not the comparison operator "==" Link to comment https://forums.phpfreaks.com/topic/107265-solved-quick-database-question/#findComment-550085 Share on other sites More sharing options...
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