big-dog1965 Posted March 7, 2009 Share Posted March 7, 2009 Can someone show me what is wrong here line 15 is $result = mysql_query("Select * From login_table where user_name='$username'; Parse error: syntax error, unexpected ')' in /home/public_html/App/Admin/login.php on line 15 if (isset($_POST['Submit'])) { $username = $_POST['username']; $password = md5($_POST[password]); mysql_connect($dbServer, $dbUsername, $dbPassword)){ $result = mysql_query("Select * From login_table where user_name='$username'; if(mysql_num_rows($result)>0) { $row = mysql_fetch_array($result, MYSQL_BOTH); if($password == $row["user_pass"]) { $_SESSION['loginok'] = "ok"; $_SESSION['username'] = "username"; $_SESSION['password'] = "password"; $_SESSION['level'] = $row["user_level"]; Link to comment https://forums.phpfreaks.com/topic/148329-sql-connection-and-select-error/ Share on other sites More sharing options...
corbin Posted March 7, 2009 Share Posted March 7, 2009 mysql_query("Select * From login_table where user_name='$username'; Should be mysql_query("Select * From login_table where user_name='$username'"); And mysql_connect($dbServer, $dbUsername, $dbPassword)){ That's not valid syntax. Link to comment https://forums.phpfreaks.com/topic/148329-sql-connection-and-select-error/#findComment-778736 Share on other sites More sharing options...
big-dog1965 Posted March 7, 2009 Author Share Posted March 7, 2009 What should it be mysql_connect($dbServer, $dbUsername, $dbPassword)){ Link to comment https://forums.phpfreaks.com/topic/148329-sql-connection-and-select-error/#findComment-778749 Share on other sites More sharing options...
Philip Posted March 7, 2009 Share Posted March 7, 2009 $link = mysql_connect($host, $user, '$password); no { needed, just a ; Link to comment https://forums.phpfreaks.com/topic/148329-sql-connection-and-select-error/#findComment-778754 Share on other sites More sharing options...
big-dog1965 Posted March 7, 2009 Author Share Posted March 7, 2009 seems thats fix but now I get this Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /home/public_html/App/Admin/login.php on line 18 ***if(mysql_num_rows($result)>0) if (isset($_POST['Submit'])) { $username = $_POST['username']; $password = md5($_POST[password]); mysql_connect($dbServer, $dbUsername, $dbPassword); $result = mysql_query("Select * From login_table where user_name='$username'"); if(mysql_num_rows($result)>0) { $row = mysql_fetch_array($result, MYSQL_BOTH); if($password == $row["user_pass"]) { $_SESSION['loginok'] = "ok"; $_SESSION['username'] = "username"; $_SESSION['password'] = "password"; $_SESSION['level'] = $row["user_level"]; Link to comment https://forums.phpfreaks.com/topic/148329-sql-connection-and-select-error/#findComment-778773 Share on other sites More sharing options...
Luodeni Posted March 7, 2009 Share Posted March 7, 2009 seems thats fix but now I get this Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /home/public_html/App/Admin/login.php on line 18 ***if(mysql_num_rows($result)>0) if (isset($_POST['Submit'])) { $username = $_POST['username']; $password = md5($_POST[password]); mysql_connect($dbServer, $dbUsername, $dbPassword); $result = mysql_query("Select * From login_table where user_name='$username'"); if(mysql_num_rows($result)>0) { $row = mysql_fetch_array($result, MYSQL_BOTH); if($password == $row["user_pass"]) { $_SESSION['loginok'] = "ok"; $_SESSION['username'] = "username"; $_SESSION['password'] = "password"; $_SESSION['level'] = $row["user_level"]; if(mysql_num_rows($result)>0) since it's for a user try this: if ( mysql_num_rows ($result) == 1 ) OR if ( mysql_num_rows ($result) != 0 ) Link to comment https://forums.phpfreaks.com/topic/148329-sql-connection-and-select-error/#findComment-778788 Share on other sites More sharing options...
sasa Posted March 7, 2009 Share Posted March 7, 2009 change $result = mysql_query("Select * From login_table where user_name='$username'"); to $result = mysql_query("Select * From login_table where user_name='$username'") or die(mysql_error()); Link to comment https://forums.phpfreaks.com/topic/148329-sql-connection-and-select-error/#findComment-778796 Share on other sites More sharing options...
alphanumetrix Posted March 7, 2009 Share Posted March 7, 2009 The problem is simple. You forgot to close it. This will work: $result = mysql_query("Select * From login_table where user_name='$username'"); You forgot to finish the statement with "); instead, you just inded it with '; Link to comment https://forums.phpfreaks.com/topic/148329-sql-connection-and-select-error/#findComment-778843 Share on other sites More sharing options...
big-dog1965 Posted March 7, 2009 Author Share Posted March 7, 2009 change $result = mysql_query("Select * From login_table where user_name='$username'"); to $result = mysql_query("Select * From login_table where user_name='$username'") or die(mysql_error()); I get error could not select database. so from the above code I added mysql_connect($dbServer, $dbUsername, $dbPassword, $dbDatabase); but still didnt work Link to comment https://forums.phpfreaks.com/topic/148329-sql-connection-and-select-error/#findComment-778938 Share on other sites More sharing options...
redarrow Posted March 7, 2009 Share Posted March 7, 2009 try a diffrent database connection code as your's is all flash. <?php $db=mysql_connect("localhost","username","password"); $res=mysql_select_db("database_name",$db)or die('database connection error'.mysql_error()); ?> You was crawling before walking when you grow up, i guess. Never go above your head, unless you no what your doing. Link to comment https://forums.phpfreaks.com/topic/148329-sql-connection-and-select-error/#findComment-778956 Share on other sites More sharing options...
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