BrandonLayton Posted December 7, 2010 Share Posted December 7, 2010 For some reason this code is not changing anything in the database. I am 99% sure I am typing in the right info into the text boxes, and even if I wasn't it should give me error messages. Kinda stumped. Hopefully a new pair of eyes can help. <!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>Member System Install</title> </head> <body> <form id="form1" name="form1" method="post" action=""> <h2><center>Fill in the following info properly to install required databases.</center></h2> <p> <label>Host: <input type="text" name="host" id="host" /> </label> </p> <p> <label>Database User: <input type="text" name="dbUser" id="dbUser" /> </label> </p> <p> <label>Database Password: <input type="text" name="dbPass" id="dbPass" /> </label> </p> <p> <label>Desired Admin Username: <input type="text" name="adminName" id="adminName" /> </label> </p> <p> <label>Desired Admin Password: <input type="text" name="adminPass" id="adminPass" /> </label> </p> <p> <label>Email: <input type="text" name="adminEmail" id="adminEmail" /> </label> </p> <p> <input type="submit" name="submitBtn" id="submitBtn" value="Submit" /> </p> </form> <?php if(isset($_POST['submit'])){ $hostName = $_POST['host']; $dbUserName = $_POST['dbUser']; $dbPassword = $_POST['dbPass']; $adminUserName = $_POST['adminName']; $adminPassword = $_POST['adminPass']; $email = $_POST['adminEmail']; $sqlLink = mysql_connect('$hostName','$dbUserName','$dbPassword'); if(!$sqlLink){ die('Could not connect: ' .mysql_error()); mysql_close(); } else { $queryCreate = 'CREATE DATABASE member_db'; echo 'Connected successfully!'; if(mysql_query($queryCreate,$link)){ echo 'Created database!'; mysql_select_db('member_db'); $createAdminTable = "CREATE TABLE " .$adminUserName. " (`secLvl` tinyint(1) default NOT NULL,`rank` tinyint(2) default NOT NULL,`username` varChar(32) NOT NULL UNIQUE,`password` varChar(32) NOT NULL,`email` varChar(32) NOT NULL UNIQUE"; $insertAdminInfoQuery = "UPDATE " .$adminUserName. " SET `secLvl` = '5', `rank` = '10', `username` = '$adminUserName', `password` = '$adminPassword', `email` = '$email' WHERE '$adminUserName' LIMIT 1"; mysql_query($createAdminTable) or die("ERROR: " .mysql_error()); mysql_query($insertAdminInfoQuery) or die("ERROR: " .mysql_error()); } else { echo 'Error: '.mysql_error(); } } } ?> </body> </html> Thanks, Brandon Quote Link to comment Share on other sites More sharing options...
KevinM1 Posted December 8, 2010 Share Posted December 8, 2010 You need: if(isset($_POST['submitBtn'])) Also, you've got your logic backwards. With PHP, you always attempt to handle incoming POST data first, then show the form. Your form should still send the data with the way you have it now, but you're coding yourself into a corner with your current script structure. Quote Link to comment Share on other sites More sharing options...
BrandonLayton Posted December 8, 2010 Author Share Posted December 8, 2010 Wow. gosh I need to check stuff like that a little better. Thanks Ok, only used php a few times so not advanced enough to really know that I guess xD Thanks a lot Quote Link to comment Share on other sites More sharing options...
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