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error in your SQL syntax...ear 'Resource id #3' ???

P: 4
i have this query:
$query = mysql_query("SELECT F_ID,F_TITLE,A_FNAME,A_LNAME FROM (take_part INNER JOIN film ON F_ID = T_P_ID)INNER JOIN actor ON A_PID = ACTOR_ID WHERE A_PID=\"$a_id\"");
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'Resource id #3' at line 1
Jul 8 '06 #1
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1 Reply

P: 4
plese help me....
thanks in advanced!
Jul 8 '06 #2

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