PHP and 2x connection to DB

czezz · 02-11-2012, 03:38 PM

In PHP code I open 2 connection to 2 different databases (on different hosts).

PHP Code:


			
$dbh1 = mysql_connect($hostname, $username, $password) or die(mysql_error()); 
mysql_select_db('database1', $dbh1);
mysql_query('select * from tablename', $dbh1);

/*
and for database 2
*/

$dbh2 = mysql_connect($hostname, $username, $password)  or die(mysql_error());
mysql_select_db('database2', $dbh2);
mysql_query('select * from tablename', $dbh2);

That works just fine.
However, if connection to first DB - $dbh1 - fails, then I can see only error and no data form $dbh2.

Actually nothing what is below $dbh1 is displayed. Even test message like: print "test";

Why is it like this and what to do to make it works?
My system details:

Code:

Apache/2.2.17 (Unix) DAV/2 PHP/5.3.6 
mysql  Ver 14.14 Distrib 5.1.56, for slackware-linux-gnu (i486) using readline 5.1

Dark_Helmet · 02-11-2012, 04:08 PM

The reason you see nothing else is because you told PHP to not do anything else.

Your command:

Code:

$dbh1 = mysql_connect($hostname, $username, $password) or die(mysql_error());

So, if the mysql_connect() fails, then the die() function is called--but ONLY if the mysql_connect() fails.

The documentation for die() states that it is equivalent to exit(). And the documentation for exit() says that the script will terminate after such a call.

If you want to continue your script regardless of whether either mysql_connect() function fails, then you need to remove the "or die()" portions and insert checks to make sure you have a valid, working connection before you submit queries.

czezz · 02-11-2012, 04:39 PM

This is correct. When I remove die() it works.
And I have added validation:

PHP Code:


			
$dbh1 = mysql_connect($hostname, $username, $password)

if(empty($dbh1))
  {
   print $c_ekt;
   print ("Can not establish connection to DB " . mysql_error());
  }

TY for reply.