Re: [PHP] opening table 2 inside table1

Curt, Thanks. > > #mysql_select_db($db2,$myconnection2); #not necessary > > Why isn't this necessary? if you have just opened a connection you > must select a database. Besides the fact that the second connection > isn't necessary. > >From what I have experimented with, if I declare $db.$table in

Re: [PHP] opening table 2 inside table1

* Thus wrote John Taylor-Johnston: > I want to open table 2 inside table1. See $sql2 below. I know this is illegal. what your doing is perfectly legal. > How can I work around this? Or can I? Lost cause? I'm not sure what kind of behaviour your getting but I'm going to blow through your code poi

[PHP] opening table 2 inside table1

I want to open table 2 inside table1. See $sql2 below. I know this is illegal. How can I work around this? Or can I? Lost cause? John $myconnection = mysql_connect($server,$user,$pass); #mysql_select_db($db,$myconnection); #not necessary $sql1 = 'select * from '.$db.'.'.$table.' order by district,