You're running 3 entirely independent scripts, so they can't share a connection.

You may save an infinitessimal amount of time by using mysql_pconnect() instead of 
mysql_connect(), which will give each process/thread a permanent database connection, 
so it doesn't have to connect every time.

Adam Hooper
[EMAIL PROTECTED]

On Fri, 31 May 2002 09:55:27 +0100
"Caspar Kennerdale" <[EMAIL PROTECTED]> wrote:

> Sorry If this is the wrong list for this topic, I hope someone can shed some
> light onto my problem.
> 
> I am building a php/ Mysql web site for a client which is a picture gallery.
> The web site has 3 frames (required by the designer so that the whole thing
> doesnt refresh all the time).
> Frame 1- Navigation
> Frame 2- Info on selected artworks and other related projects
> Frame 3- The artworks/ jpegs
> 
> I have one table, with a name and info about the project and upto 5 urls of
> where the jpeg lies.
> 
> When a project is selected in the navigation I request the record from the
> database, I create an array which contains the location of the jpegs and
> then display them in Frame 3.
> 
> Now, I then have a piece of javascript which tells frame 2 to update itself.
> So I have parsed the record ID to it and it then open a query to the
> database and outputs the relevant information about the artworks. Lastly a
> third query is also sent to the database to see if there are any other
> projects in the gallery by the same artist- and then create a list of
> related links.
> 
> So I have 3 database queries over two pages.
> 
> I'm wondering if there is a more efficient way of doing this?
> 
> 
> 
> 
> 
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