grifb
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Warning: mysql_num_rows() expects parameter 1 to be resource
grifb replied to grifb's topic in PHP Coding Help
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Warning: mysql_num_rows() expects parameter 1 to be resource
grifb posted a topic in PHP Coding Help
help!!! i have been trying things all day with this and it wont work! <html> <div align="center"> <?php $key = $_POST['key']; $answer = $_POST['answer']; $username = $_POST['username2']; $item = explode(" ", $key); $final = $item[0]; $removals = array("a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z"); $lower = str_replace($removals, "", $final); $newkey = strtolower($lower); $areplace = str_replace("," , "', '", $answer); $buff = "['$areplace']"; $finalanswer = strtolower($buff); if ($key&&$username&&$answer) { $connect = mysql_connect("fakedatabase","fakeuser", "fakepassword") or die("Couldn't Connect"); mysql_select_db("wordlist") or die("No Table"); $query = mysql_query("SELECT * FROM keylist WHERE keys='$newkey'"); $numrows = mysql_num_rows($query); echo $numrows; if ($numrows!=0) { $queryadduser1 = mysql_query("INSERT INTO userkeys Values('$username', 'Tried : $newkey')") or die("Unable to connect"); die("The key '$newkey' has already been added!"); } else $queryadd = mysql_query("INSERT INTO keylist Values('$newkey')") or die("Unable to connect"); $queryadd2 = mysql_query("INSERT INTO KeysAndAnswers Values('$newkey', '$finalanswer')") or die("Unable to connect"); $queryadduser1 = mysql_query("INSERT INTO userkeys Values('$username', 'Added : $newkey ')") or die("Unable to connect"); echo("The key '$newkey' and the answer '$finalanswer' Has just been submitted! Thank you!"); } else die("Key or answer or username was blank"); ?> </div> </html> You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'keys='123-123'' at line 1