RE: [PHP] What am I doing wrong here?

$result = mysql_query("SELECT * FROM users WHERE first_name=$first_name AND last_name=$last_name"); Try $result = mysql_query("SELECT * FROM users WHERE first_name='$first_name' AND last_name='$last_name'"); Cal -----Original Message----- From: Lynn Siprelle [mailto:lynn@siprelle.com] Sent: Monday, December 25, 2000 7:01 PM To: php-general@lists.php.net 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