is a nice solution of it.$db1 = mysql_connect("hostname","username","password")
mysql_select_db(”db1″, $db1);
$res1 = mysql_query("query",$db1);
$db2 = mysql_connect("hostname","username","password");
mysql_select_db("db2″, $db2);
$res2 = mysql_query("query",$db2);
At this point you can only fetch records from you previous ResultSet,
i.e $res1 – But you cannot execute new query in $db1, even if you
supply the link as because the link was overwritten by the new db, So at this point the following script will fail
$res3 = mysql_query("query",$db1); //this will failSo how to solve that?
take a look below.
$db1 = mysql_connect("hostname","username","password");
mysql_select_db("db1″, $db1);
$res1 = mysql_query("query",$db1);
$db2 = mysql_connect("hostname","username","password", true)
mysql_select_db("db2", $db2);
$res2 = mysql_query("query",$db2);
So mysql_connect has another optional boolean parameter which
indicates whether a link will be created or not. as we connect to the
$db2 with this optional parameter set to ‘true’, so both link will
remain live.
now the following query will execute successfully.
$res3 = mysql_query("query",$db1);
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