I wrote this little snippet of code to display my average posts per day on my journal. Unfortunately that number turned out embarrassingly small (0.383) so I added functionality to display how many days between posts (a post every 2 days looks better). Just uncomment the echo you prefer.
Requires PHP and MySQL.
1 | Arvind
Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /home/arvinds/public_html/blog/sidebar2.php on line 298
Warning: Division by zero in /home/arvinds/public_html/blog/sidebar2.php on line 307
Why does this happen. Edited the db stuff and changed date of first post
3 | KMB
Hi there
I think it could be a bit faster when you change the SQL-Statement to something like this:
"SELECT id FROM TABLE"
Now MySQL don't have to get all the info from the whole table. You probably have to change id to the name of a column in your table. Don't know if MT is using IDs…
Btw: Arvind you have to change the name of the TABLE too.
4 | Michelle
Right, you have to change USERNAME, PASSWORD, DATABASE, and TABLE. I did it that way because I use my own journaling script rather than Moveable Type, and I don't know what the table is called, neither do I know if MT uses a column called ID. This way it's pretty much compatible with anything that uses a MySQL db.
6 | Arvind
Is there a way to restrict the amount of dp (decimal places) its given to, mine is avg. one post every 1.6266666666667 days which is a bit annoying 😉
7 | Michelle
Yeah, it took me a minute, but add this just before the echo:
$posts_per_day = substr ( $posts_per_day, 0, ### );
$posts_per_day = substr ( $posts_per_day, 0, strlen( $posts_per_day ) – strpos( strrev( $posts_per_day ), " " ) );
$posts_per_day = trim ( $posts_per_day );
$days_per_post = substr ( $days_per_post, 0, ### );
$days_per_post = substr ( $days_per_post, 0, strlen( $days_per_post ) – strpos( strrev( $days_per_post ), " " ) );
$days_per_post = trim ( $days_per_post );
Replace both of the ### with the number of characters you want shown, INCLUDING the decimal. So 1.62 would be 4 characters, 1.6 would be 3, etc.
8 | Michelle
Actually, that has an extra line in it, that makes it not end in the middle of a word, which isn't necessary here. You can just add this (and replace the ### both times):
$posts_per_day = substr ( $posts_per_day, 0, ### );
$posts_per_day = trim ( $posts_per_day );
$days_per_post = substr ( $days_per_post, 0, ### );
$days_per_post = trim ( $days_per_post );
9 | Michelle
I should do more testing before I post… It turns out you only need the first line for each variable:
$posts_per_day = substr ( $posts_per_day, 0, ### );
$days_per_post = substr ( $days_per_post, 0, ### );
I copied this straight from the rss feed I wrote, and I guess I didn't understand the concept as well as I thought. Hey, at least my feed works, right? 😉
12 | mintaboo
not to split hairs, but
select * from dbo.table; is extremely inefficent. say you have 10 fields in the table, at least two contain large portions of text, and then have 3,000 entries (i.e. a blog table). i would say each time you called that little code snippet, you've requested 25 or more megabytes of information (infact you'd have enough info in the server's RAM to display every blog entry you've ever written!). multiply this by the number of simulanteous users (say 5).. thats 125 megs of data being sloshed around and you haven't even retrieved the data for the rest of the page! 
thats a pretty conservative estimate at that. whats the solution? well, i personally would use an aggregate function, namely COUNT.
select COUNT(blog_id) from blogtable;
the first column in the first row will contain the desired number. avoid COUNT(*) as that will replicate the aforementioned '*' issue.
good look, and godspeed.
mint
13 | Code Novice
14 | Code Novice
scriptygoddess
