note 57533 deleted from function.mysql-connect by bjori

Note Submitter: baldurien at club-internet dot fr ---- There are some case where you will have one user, for example root, that may use different database, for example : <?php $x = mysql_connect( ... ); mysql_select_db('foo', $x); /* some queries on foo */ mysql_select_db('bar', $x ); /* some queries on bar */ ?> This is fine in that case, however if you would like to transfer one table from database "foo" to database "bar", you may do that : INSERT INTO bar.desttable SELECT * FROM foo.srctable (or CREATE TABLE) But if you would like to convert data using a specific function, then you will rather use that : <?php $x = mysql_connect( 'localhost', 'root', 'root' ); $y = mysql_connect( 'localhost', 'root', 'root' ); mysql_select_db('foo', $x); mysql_select_db('bar', $y ); $q = mysql_query( 'SELECT field1, ..., fieldN FROM srctable', $x ); while ( $a = mysql_fetch_row($q)) mysql_query( 'INSERT INTO bar VALUES(\" . foo($a[0]) . '\', ..., \'' . foo($a[0]) . '\'', $y); ?> This will fail. The reason is simple : PHP will reuse the existant connection for the second mysql_connect, so when you will select a database on server $x, you will also select it for $y, and vice versa! This is why you have to use the fourth parameter to avoid this : <?php $x = mysql_connect( 'localhost', 'root', 'root' ); $y = mysql_connect( 'localhost', 'root', 'root', true ); ?> This way your script will work. Note : doing that mysql_close on $x will result in closing the connection for $y too. <?php $x = mysql_connect( 'localhost', 'root', 'root' ); $y = mysql_connect( 'localhost', 'root', 'root' ); mysql_close($x); mysql_close($y); // server is already closed ?>