I am working on an application that requires me to connect to two
MySQL databases both hosted on the same server.
I have an existing set of class files I created to handle the db
connection to the main database. To connect to the second database,
I duplicated my db class files and renamed them, and also renamed
the functions etc. but what is happening, is that the calls made to
the second database connection seem to get made against the first
database. So that I get errors like: dbname.tablename was not found
(or something like that).
What is the best practice for this? I need to have two open db
connections for just two little things in my application.
In my opinion, rather than making two different classes that do the
same thing, I would make one class and then create two instances of
that class in your script. For this to work, you'll need to make the
db host, username, password, and database name be values that are
passed to the constructor when the object is created. Assuming you
have that, then your code would be something like...
$dbc1 = new dbconn('localhost', 'test', 'test', 'mydb');
$dbc2 = new dbconn('localhost', 'test', 'test', 'otherdb');
Then invocations of $dbc1->retrieveData() would to go the one
database whereas $dbc2->retrieveData() would go to the other. I could
be missing something, of course, but that's my first thought.
Also, just to be safe, you could use the new link argument in
mysql_connect() (the 4th argument) to insure a new, separate connection.
Hope that helps,
Larry
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