I'm not sure if this was addressed before. Couldn't find it by searching the forum.
I was wondering if someone can tell me how to develop globals to display:
1. Total number of reviews in the directory
2. Total number of links with reviews

Thanks.
----------------

waffle

Total number of reviews is fairly easy,

sub {
return ( $DB->table ('Reviews')->total )
}

Getting the count is trickier, since the 'do' method was apparantly removed. The SQL statement
is

SELECT COUNT(DISTINCT Review_LinkID) FROM Reviews

But I can't figure out how to get it into a method call.


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I *think* this would work for number 2...

$DB->table('Links', 'Reviews')->count;

It will use the foreign key to do the count and so should return the number of links with reviews.

Alternatively this should work....

my $total = $DB->table('Reviews')->('DISTINCT(Review_LinkID)')->fetchall_list;

$DB->table('Links', 'Reviews')->count;

Wouldn't that return the total number of reviews rather than the total number of links with reviews? I assume that count just counts the number of rows in the joined table which should have multiple rows per link?

I use rows like this but I'm not sure what the difference is.

my $total = $DB->table('Reviews')->('DISTINCT Review_LinkID')->rows;

I think the resulting query is something like:

SELECT COUNT(*) FROM Links,Reviews WHERE Links.ID = Reviews.Review_LinkID

So it will count the number of links that have a corresponding review and hence the number of reviews.

I've not tested so I may be talking balls =)

I agree with your query but still think that this will give the number of reviews?

Say you have 1 link and it has 5 reviews. Wouldn't the joined table have 5 rows?

Yeah I just tested - you're right.

My revised code :)

What was wrong with my code? Don't see why you need the table join?

my $total = $DB->table('Reviews')->('DISTINCT Review_LinkID')->rows;

Good point - it's late......(that's my excuse)

Oh one thing wrong with your code is you are missing a function name

Good point - whoops.