$query = sprintf("INSERT INTO people (userName, userAddress, userCity, userZip, userEmail) VALUES( '%s', '%s','%s','%s','%s')", PHP: What are those %s ?
you are missing the last part of the code... sprintf will replace those %s (strings) with values you provide. something like $string = sprintf("INSERT %s %s","string","here"); would output $string = "INSERT string here";
it's just replacing %s with some other variable that you supply. chances are you need to supply the name address zip and stuff that is being inserted into the table