Mavrik347

Hehe I know that now

This is the query PHPMyAdmin generated: SELECT * FROM `users` WHERE `tsName` LIKE '%IRNP%'; I was not aware however that there was any difference between a ` and a '.

SELECT * FROM users WHERE tsName LIKE '$cTicker%'; Thanks man, so much. Had my head in a spin there for a minute. I generally know SQL syntax, the reason I was so confused is that the original query I typed into PHPMyAdmin worked straight off the bat. Then when I made it generate the query for me it generated the same thing I just typed. So I thought it must be right. All working as you can see. Thank you lovely chaps

lol care to throw me a bone?

Thanks for the fast reply! I tried both SELECT * FROM 'users' WHERE tsName 'LIKE $cTicker%'; SELECT * FROM 'users' WHERE tsName 'LIKE' $cTicker%; SELECT * FROM 'users' WHERE tsName LIKE% $cTicker; SELECT * FROM 'users' WHERE tsName 'LIKE%' $cTicker; SELECT * FROM 'users' WHERE tsName LIKE '$cTicker%'; But still the same error. :/ You're right though, it was very silly of me to not echo the mysql_error() "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 ''users' WHERE tsName 'LIKE%' A-43' at line 1"

Hello guys, I hope you can help. I can't get my page to switch between two database connections properly It seems to ignore the second connection. I'm trying to go through each group in database one and find out how many members they have registered in database two by checking the field "tsName". This is a valid query for database one. $fetchUsers = mysql_query("SELECT * FROM $db_corps_table ORDER BY cName;", $con1); This is a valid query for database two. $fetchUsers = mysql_query("SELECT * FROM 'users' WHERE tsName LIKE $cTicker%;", $con2); However despite me specifying which database link to use it tries to use every query on database one ($con1). So I can never query database two ($con2) because it just breaks with the result below. :S $fetchUsers = mysql_query("SELECT * FROM $db_corps_table ORDER BY cName;", $con1); gives But: $fetchUsers = mysql_query("SELECT * FROM 'users' WHERE tsName LIKE $cTicker%;", $con2); gives <?php // ===== Connect to database one (group) ===== $con1 = mysql_connect($db_host,$db_corps_user,$db_corps_pass); if (!$con1) { die("Could not connect:" . mysql_error()); } $db_select1 = mysql_select_db($db_corps_db, $con1); if (!$db_select1) { die ("Could not select database:" . mysql_error()); } // ===== Connect to database two (members) ===== $con2 = mysql_connect($db_host,$db_ts_user,$db_ts_pass, true); if (!$con2) { die("Could not connect:" . mysql_error()); } $db_select2 = mysql_select_db($db_ts_db, $con2); if (!$db_select2) { die ("Could not select database:" . mysql_error()); } $fetchCorps = mysql_query("SELECT * FROM $db_corps_table ORDER BY cName;", $con1); while ($row = mysql_fetch_array($fetchCorps)) { // For each of the results, gather details from database one, then see how many members we can find in database two $cName = $row["cName"]; $cTicker = mysql_real_escape_string($row["cTicker"],$con1); $cMembers = $row["cMembers"]; echo $cName." [".$cTicker."] (".$cMembers."):<br />"; // ========================== HERE BE PROBLEMS ========================== //$fetchUsers = mysql_query("SELECT * FROM $db_corps_table ORDER BY cName;", $con1); $fetchUsers = mysql_query("SELECT * FROM 'users' WHERE tsName LIKE $cTicker%;", $con2); // =================================================================== while ($rowTS = mysql_fetch_array($fetchUsers)) { echo "result<br />"; } } // all done, close DB connections mysql_close($con2); mysql_close($con1); ?>