How could I display total # of subscribers? Help Please

[TrickyD]

I use the paid subscriptions feature to help cover a majority of the financial burden my forum generates and I'd like to be able to display somewhere in the header the total number of subscribers.

Basically I'd like to have just a small little space say

"Current Subscribers X, Goal 120"

Where X is pulled automatically like it is from the Subscriptions Manager tab in the AdminCP

If anyone can help on this one I'd greatly appreciate it. Thanks!

[TrickyD]

Can any of you code geniuses out there help me out with this?

[Andreas]

[sql]select count(*) from subscriptionlog where status=1[/sql]

[TrickyD]

Originally Posted by KirbyDE

[sql]select count(*) from subscriptionlog where active=1[/sql]

Kirby, thanks for the reply.

How will running that query allow me to call the value in the header template?

[Andreas]

Well, just execute it

global_start

PHP Code:

$subscribers = $db->query_first("SELECT COUNT(*) AS totalactive FROM " . TABLE_PREFIX . "subscriptionlog WHERE active=1");
$subscribers['totalactive'] = vb_number_format($subscribers['totalactive']); 


As this adds 1 Query/Page, you might want to cache the Value in Datastore, but that involves more Code - and File Edits.

[TrickyD]

Originally Posted by KirbyDE

Well, just execute it

global_start

PHP Code:

$subscribers = $db->query_first("SELECT COUNT(*) AS totalactive FROM " . TABLE_PREFIX . "subscriptionlog WHERE active=1");
$subscribers['totalactive'] = vb_number_format($subscribers['totalactive']); 


As this adds 1 Query/Page, you might want to cache the Value in Datastore, but that involves more Code - and File Edits.

Kirby, thanks again for the help, but I think I'm code stupid.

I attempt to run the query but it gives me the following error

An error occurred while attempting to execute your query. The following information was returned.
error number: 1146
error desc: Table 'trickyd_ncaa.subscriptionlog' doesn't exist

And when I create a new plugin for the global_start.php as you suggested my board gets a fatal error.

[Marco van Herwaarden]

Then you are probably using a table prefix, and will have to add that in front of tablenames.

[TrickyD]

Originally Posted by MarcoH64

Then you are probably using a table prefix, and will have to add that in front of tablenames.

Indeed, my fault there. However when adding the prefix to the query, it returned the following error.

An error occurred while attempting to execute your query. The following information was returned.
error number: 1054
error desc: Unknown column 'active' in 'where clause'

Any thoughts?

[Marco van Herwaarden]

What is the exact query you are running?

If you are using a table prefix, and are referencing the tablename as identifier for a columnname anywhere the SQL-statement, you should also add an AS clause. Example:

PHP Code:

SELECT table1.name, table2.*
FROM " . TABLE_PREFIX . "table1 AS table1
LEFT JOIN " . TABLE_PREFIX . "table2 AS table2 ON (table1.keyid = table2.keyid) 


[Andreas]

My fault - it's status not active