VARIABLE IN SQL STATEMENT! QUICK!!!!!!

[Anon]

Need help with this fast!

I AM TRYING TO PASS THE DECLARED VARIABLE ($pick_number) INTO A SQL STATEMENT BELOW:

$query = "SELECT * FROM $brackettable where (username = '$bracket_username' AND password = '$bracket_password' AND $pick_number LIKE '%')";

DISPLAYS AS:

$query = "SELECT * FROM $brackettable where (username = '$bracket_username' AND password = '$bracket_password' AND LIKE '%')";

$pick_number (although already declared) does not show...WHY????????????

PLEASE HELP!!!!!!!!!!!!!!!!!!!

[Anon]

$query = "SELECT * FROM $brackettable where (username = '$bracket_username' AND password = '$bracket_password' AND $pick_number LIKE '%')";

should be

$query = "SELECT * FROM $brackettable where (username = '$bracket_username' AND password = '$bracket_password' AND '$pick_number' LIKE '%')";

hope this helps!

[Anon]

THAT DOESNT WORK EITHER...TRIED THAT. IT GIVES ME THIS: ''
WITH NOTHING IN BETWEEN THE SINGLE QUOTES?

I AM POSITIVE THE VARIABLE HAS A VALUE

[Anon]

GOT IT...
YOU WERE RIGHT!

I put the variable into the "IF" statement rather than outside of it and it worked...with single quotes

thanks again

[Anon]

NP, I was about to ask you to check where you declared the values next - 😉

Regards,
Charlie aka Chuck