Parse error: syntax error, unexpected T_STRING for one line but not the other.

Simplyput

I am not sure if it's me or some new set up they have on my server. I have the following line below where I get this error message.


$inactive = $db->sql_fetchrowset($db->sql_query (("SELECT COUNT(mid)FROM pol_active_starter INNER JOIN pol_pbp_starter ON (pol_active_starter.'phone_number' = pol_pbp_starter.'verrus_lot_id')WHERE pol_active_starter.'phone_number' = pol_pbp_starter.'verrus_lot_id' AND pol_pbp_starter.'date_start' != ".$today.")));

$today is defined as $today = (Date("l F d, Y"));

However, I have a line above it with the following and I get no error message.


$active = $db->sql_fetchrowset($db->sql_query (("SELECT COUNT(mid)FROM pol_active_starter INNER JOIN pol_pbp_starter ON (pol_active_starter.'phone_number' = pol_pbp_starter.'verrus_lot_id')WHERE pol_active_starter.'phone_number' = pol_pbp_starter.'verrus_lot_id' AND pol_pbp_starter.'date_start' = ".$today.")));

$today is defined as $today = (Date("l F d, Y"));

What the badonkey dunk and I doing wrong. I have been recoding and searching the internet for solutions for two days now and nothing seams to work. Please help.

Derokorian

The first one causes the problem, however the parser doesn't notice the problem until later. The problem comes the fact you reopen the string after the $today variable, and then that the today variable is not a string in SQL. Notice how in the syntax highlighting above, your closing parens are all red instead of green like the opening parens? Any good IDE/Editor would show you the same thing, but it means those parens are in a string, and not closing any function calls as you might expect.

sneakyimp

Having a good editor or Integrated Development Environment (IDE) helps a lot for catching this kind of bug. Just to repest Derokorian, you failed to provide a closing double quote after $today."))); and PHP thinks what comes after that is all part of the same string until it encounters the next double quote. It then chokes on the text that follows that next double quote (which is really just the beginning of another sql statement) because the SQL string that follows the second double quote doesn't parse like valid PHP.