Re: What am I doing wrong here?

Try this and see what happens: $result = mysql_query("SELECT * FROM users WHERE first_name=$first_name AND last_name=$last_name") or die("Error on " . __LINE__ . " of " . __FILE__ . ": " . mysql_error()); Typically if it doesn't see $result as a valid resource it means your query failed; if that's the case, adding the or die(...) to it will tell you why it's failing ;) Toby Butzon [ criticism spurs improvement ] ----- Original Message ----- From: "Lynn Siprelle" <lynn@siprelle.com> To: <php-general@lists.php.net> Sent: Monday, December 25, 2000 8:00 PM Subject: [PHP] What am I doing wrong here? : Hi guys-- : : I'm going through a couple of tutorials on user validation/tracking, and I come up with the same kind of problem over and over again. (Background: MySQL, PHP4, FreeBSD, Apache w/mod_php, full permissions to do whatever I'd like pretty much. Have verified that PHP and MySQL are working fine otherwise. I am a newbie, however.) : : Here's the problem: Whenever I'm trying to verify whether someone has already entered a user name into the database, I get some kind of error. Here's an example taken from a devshed tutorial of a script that gives me this trouble (and I do have the username, password and database correctly entered), using an html form page to submit a first name and a last name into the database via this script: : : <html> : <body> : <?php : : mysql_connect (localhost, myusername, mypwd); : : mysql_select_db (mydb); : : $result = mysql_query("SELECT * FROM users WHERE first_name=$first_name AND last_name=$last_name"); : : if (mysql_num_rows($result) == 1){ : print "Sorry, someone has already entered your name into the database."; : } : else { : mysql_query ("INSERT INTO users (first_name, last_name) VALUES ('$first_name', '$last_name')"); : : print ($first_name); : : print (" "); : : print ($last_name); : : print ("<p>"); : : print ("Thanks for submitting your name."); : } : : ?> : </body> : </html> : : This particular version of trying for verification produces the error : : Warning: Supplied argument is not a valid MySQL result resource in /path/to/submitform.php on line 11 : : And then it goes ahead and executes the "else" statement, even if the name already exists in the database; when I go into MySQL and check, there are now two entries of my test name. I have tried at least one other tutorial on this with similar results (a parse error instead of a "not a valid MySQL result resource" error) and have dug through the mysql functions in the manual trying to see what I'm doing wrong (or at least figure out why I can't get a pre-scripted tutorial to run). I should add that I believe the tutorials I'm working through are geared to php3 if I'm not mistaken, but I thought php4 was backwards compatible, nu? I should also add that the script executes fine WITHOUT the validation section, neatly placing the name into the table and echoing back a polite thank-you. : : Ideas? I'm a newbie, so it very well may be something stoopid I'm doing; be gentle but firm. :) : : Lynn : -- : Siprelle & Associates: Distinctive Web Design Since 1994 : Lynn Siprelle, President: Newest Associate Josie born 9/9/97 : Business: lynn@siprelle com | Personal: lynsa@siprelle.com : Portfolio: http://www.siprelle.com/ | TNH: http://www.newhomemaker.com/ : : -- : PHP General Mailing List (http://www.php.net/) : To unsubscribe, e-mail: php-general-unsubscribe@lists.php.net : For additional commands, e-mail: php-general-help@lists.php.net : To contact the list administrators, e-mail: php-list-admin@lists.php.net : :