How do you deal with need for MySQL connection inside function, outside function

[cgm225]

I have a script that connects to one database (DB #1) for use throughout that script, which I close at the end of the entire script.  However, within that script I have a function I use to connect to a second database (DB #2), which I grab some other information from.  I end that function by closing that mysql connection. 

 

However, now I am getting "invalid MySQL resource" errors from the script after the point which I use that function.  So I wanted to know, how do you deal with multiple MySQL database connections within one script, particularly in the setting I have outlined above?

 

Thanks in advance!

[mem0ri]

Personally, I wouldn't change the entire connection, only switch databases within the same connection.

 

mysql_select_db() is the function...

 

...that way...with the function you can start it with mysql_select_db($x) and then go back to mysql_select_db($y) at the end...

 

...unless you're connecting to two completely different db servers?

[wildteen88]

Make sure you provide the link resource whenever you call a mysql function which requires it. eg:

$conn1 = mysql_connect('server1', 'userr1', 'pass1); # server1
$conn2 = mysql_connect('server2', 'userr2', 'pass2');  # server2

# select databases
mysql_select_database('server1_db', $conn1); # server1 database
mysql_select_database('server2_db', $conn2); # server2 database

#query server
$server1_qry = mysql_query('Select * from tbl_name', $conn1); # query server 1 database
$server2_qry = mysql_query('Select * from tbl_name', $conn2); # query server 2 database

#close connections
mysql_close($conn1) # close server1 connection
mysql_close($conn2) # close server2 connection

Variables $conn1 and $conn2 contain the link resource, you'll notice these are passed as the second parameters to the above functions. This is how you deal with multiple connections to different servers/databases as the sametime. Note not all mysql functions require the link resource to be provided you should check the manual for which functions do.

 

 

[cgm225]

Thank you both so much!

[Barand]

If they are on the same server you can just specify the db to use inside the query

 

SELECT x, y, z FROM DB2.tablename

 

or even mix'n'match

 

SELECT a.z, b.y FROM DB1.tableA as a INNER JOIN DB2.tableB as b ON a.col = b.col

[cgm225]

Still having some issues...

 

I have attached the script I am using to load images*.  It works fine, except when I try to use a function (also attached*) to get data from a second database.  When I try to use that function, called "permission_for" the MySQL query following it's use in the main script (the query used to generate the "Next" link) fails, outputting the following error message: "Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /web/example.com/includes/php/gallery/photos.inc.php on line 165"

 

Could someone help me figure out why this is happening?

 

Thank you all again!

 

 

 

*I tried to use the [] code tags to include these items, but I was getting an error when trying to do so.  I apologize for the inconvenience of the attachments.

[cgm225]

bump