todayme Posted March 2, 2007 Share Posted March 2, 2007 I have the following lines of code I need to valuate what is in $data and if it doesnt match the string I need it to exit execution. I am a newbie any ideas? $data = mysql_query("SELECT * FROM _User WHERE Email = '$dbemail' AND Password = '$dbpassword'") or die(mysql_error()); Quote Link to comment Share on other sites More sharing options...
itsmeArry Posted March 2, 2007 Share Posted March 2, 2007 $data = mysql_query("SELECT * FROM _User WHERE Email = '$dbemail' AND Password = '$dbpassword'") or die(mysql_error()); $result = mysql_fetch_array($data); if(mysql_num_rows($result) < 1) exit(); Quote Link to comment Share on other sites More sharing options...
todayme Posted March 2, 2007 Author Share Posted March 2, 2007 $data = mysql_query("SELECT * FROM _User WHERE Email = '$dbemail' AND Password = '$dbpassword'") or die(mysql_error()); $result = mysql_fetch_array($data); if(mysql_num_rows($result) < 1) exit(); I have tried the mysql_num_rows before but it didnt like it. I just tried your code and got this error. Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /home/admin/domains/sold.au.com/public_html/test.php on line 27 Quote Link to comment Share on other sites More sharing options...
itsmeArry Posted March 2, 2007 Share Posted March 2, 2007 ooops sorry my mistake it should be mysql_num_rows($data) and not mysql_num_rows($result) Quote Link to comment Share on other sites More sharing options...
todayme Posted March 2, 2007 Author Share Posted March 2, 2007 ooops sorry my mistake it should be mysql_num_rows($data) and not mysql_num_rows($result) I have the code below it works if authentication fails, but if authentication is okay it doesnt do anything.......... do your if statments have end if in the syntax? I cant see that it does, thats what I know from VB........ I tried elseif = 1 keep going but that doesnt work either ...........any ideas.........by the way thanks heaps for help Quote Link to comment Share on other sites More sharing options...
itsmeArry Posted March 2, 2007 Share Posted March 2, 2007 you can use if(mysql_num_rows($result) < 1) exit(); else { /// your code goes here what you want to if there is record in db.... } Quote Link to comment Share on other sites More sharing options...
todayme Posted March 2, 2007 Author Share Posted March 2, 2007 you can use if(mysql_num_rows($result) < 1) exit(); else { /// your code goes here what you want to if there is record in db.... } Parse error: syntax error, unexpected T_ELSE in /home/admin/domains/sold.au.com/public_html/test.php on line 33 Sorry to be a pain but any ideas on that the code looks all good to me ........not that that means alot ha ha Quote Link to comment Share on other sites More sharing options...
itsmeArry Posted March 2, 2007 Share Posted March 2, 2007 post the code of test.php then I can tell where is the problem Quote Link to comment Share on other sites More sharing options...
todayme Posted March 2, 2007 Author Share Posted March 2, 2007 post the code of test.php then I can tell where is the problem <? //set the variables as required $dbhost = "localhost"; $dbuser = "admin_sex"; $dbpass = "sex"; $dbname = "admin_sex"; $dbemail = $_POST['email']; $dbpassword = $_POST['password']; //do not edit this, it connects the script $db = mysql_pconnect($dbhost,$dbuser,$dbpass); mysql_select_db($dbname) or die(mysql_error()); // Collects data from "friends" table //$data = mysql_query("SELECT * FROM _User") $data = mysql_query("SELECT * FROM _User WHERE Email = '$dbemail' AND Password = '$dbpassword'") or die(mysql_error()); $result = mysql_fetch_array($data); if(mysql_num_rows($data) < 1) Print "Authentication Failed"; exit(); elseif { // puts the "friends" info into the $info array $info = mysql_fetch_array( $data ); // Print out the contents of the entry Print "<b>Account ID:</b> ".$info['ID'] . " <br>"; $userid = $info['ID']; Print "<b>Name:</b> ".$info['Name'] . " "; Print "<b>Email:</b> ".$info['Email'] . " "; Print "<b>Password:</b> ".$info['Password'] . " <br>"; unset($info); Print "<br>"; Print "<br>"; // Collects data from "friends" table $data = mysql_query("SELECT * FROM _Leads WHERE LINKID = '$userid'") or die(mysql_error()); // puts the "friends" info into the $info array for($i = 0; $i < 3; $i++) { $info = mysql_fetch_array( $data ); // Print out the contents of the entry Print "<b>ID:</b> ".$info['ID'] . " "; Print "<b>Link ID:</b> ".$info['LINKID'] . " "; Print "<b>Industry:</b> ".$info['Industry'] . " "; Print "<b>Location:</b> ".$info['Location'] . " "; Print "<br>"; } unset($info); mysql_close($db); } ?> Quote Link to comment Share on other sites More sharing options...
todayme Posted March 2, 2007 Author Share Posted March 2, 2007 I thought it was else instead of elseif so I tried that but didnt work either Quote Link to comment Share on other sites More sharing options...
itsmeArry Posted March 2, 2007 Share Posted March 2, 2007 here is the error if(mysql_num_rows($data) < 1) Print "Authentication Failed"; exit(); use this if(mysql_num_rows($data) < 1) { print "Authentication Failed"; exit(); } Quote Link to comment Share on other sites More sharing options...
todayme Posted March 2, 2007 Author Share Posted March 2, 2007 Thanks dude it works well, I better get a book on php maybe php for dummies and learn the beginners syntax........ I just go off what I learned in VB and try to adapt Quote Link to comment Share on other sites More sharing options...
itsmeArry Posted March 2, 2007 Share Posted March 2, 2007 its better late than never........ Quote Link to comment Share on other sites More sharing options...
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