Re: Relational database

The info is not full but I'll try to help. Lets say the name of the field in the two tables is common_field. Then we do SELECT all_that_is_needed FROM table1,table2 WHERE table1.common_field=table2.common_field; And an optimized variant: SELECT all_that_is_needed FROM table1 LEFT JOIN table2 USING(common_field); If in the two tables we have two columns with equal data but with different names then we do: SELECT all_that_is_needed FROM table1 LEFT JOIN table2 ON table1.common_field_name_table1=table2.common_field_name_table2; So SELECT all_that_is_needed FROM table1 LEFT JOIN table2 USING(common_field); is equal to: SELECT all_that_is_needed FROM table1 LEFT JOIN table2 ON table1.common_field=table2.common_field; Hope that is clear. Best regards, Andrey ----- Original Message ----- From: "Ron" <allenr@fwd.areur.army.mil> To: <php-db@lists.php.net> Sent: Thursday, March 21, 2002 3:32 PM Subject: [PHP-DB] Relational database > I am trying to get fields from 2 different tables that have the same field > name to pull the records from one table > For example > mysql_connect($DBhost,$DBuser,$DBpass) or die("Unable toconnect to > database"); > @mysql_select_db("$DBName") or die("Unable to select database $DBName"); > $sqlquery1 = mysql_query("SELECT Description, Impact, Isolation" . > "FROM ccsd, log WHERE ccsd=log.logccsd"); > > while ($query = mysql_fetch_array($sqlquery1)) { > $CCSD = $query["CCSD"]; > $Description = $query["Description"]; > $Impact = $query["Impact"]; > $Isolation = $query["Isolation"]; > } > This code is not working.....help please.........yes I am a newbie! > > > > -- > PHP Database Mailing List (http://www.php.net/) > To unsubscribe, visit: http://www.php.net/unsub.php > >