[RESOLVED] No database selected

I have two databases and since there have to separate users for each database, I used this approach:

PHP Code:

---

$con1 = mysql_connect("www.mysite.com","user1","password") or die(mysql_error());
$con2 = mysql_connect("www.mysite.com","user2","password") or die(mysql_error());

mysql_select_db("db1",$con1) or die(mysql_error());
$aquery = "SELECT * FROM invest";
$result1= mysql_query($aquery) or die(mysql_error());
$ct1 = mysql_num_rows($result1);

mysql_select_db("microx",$con2) or die(mysql_error());
$bquery = "SELECT * FROM general";
$result2 = mysql_query($bquery) or die(mysql_error());
$ct2 = mysql_num_rows($result2); 


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The error that is showing up on the site is "No database selected".
Am I wrong in connecting to the database both at the same time or is something else wrong?

mysql_query() has an optional second argument, which is the connection the query should use. If you don't specify this argument, it'll use the last one you opened - so all of your mysql_query() calls are using $con2 (and when your first query is called, you haven't selected the database for $con2 yet, hence the error: "no database selected").

Change your calls to...
mysql_query($aquery,$con1)
mysql_query($bquery,$con2)

PHP Code:

---

// Database connection variables
$dbDatabase = "db1";
//connect to server or exit
if (!($conn = mysql_connect("www.mysite.com","user1", "password") )){
echo 'result=connection+failed';
exit;
}
//connect to database or exit
if (!(mysql_select_db($dbDatabase, $conn))){
echo 'message=db+selection+failed';
exit;
} 


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Specify the database at the top of the php file so the code knows what you are referring to.

:check:Thanks SambaNeko, that is exactly a type of solution I was looking for.