Question

I have 1 table filled with articles. For the purpose of this post, lets just say it has 4 fields. story_id, story_title, story_numyes, story_numno

Each article can be voted YES or NO. I store every rating in another table, which contains 3 fields: vote_storyid, vote_date (as a timestamp), vote_code (1 = yes, 0 = no).

So when somebody votes yes on an article, it run an update query to story_numyes+1 as well as an insert query to log the story id, date and vote_code in the 2nd table.

I would like to sort articles based on how many YES or NO votes it has. For ‘Best of all time’ rating is obviously simple…. ORDER BY story_numyes DESC.

But how would I go about doing best/worst articles today, this week, this month?

I get the timestamps to mark the cut-off dates for each period via the following:

$yesterday= strtotime('yesterday'); $last_week = strtotime('last week'); $last_month = strtotime('last month'); 

But Im not sure how to utilize these timestamps in a mysql query to achieve the desired results.

Answer 1

Try something like

SELECT id,
SUM(CASE WHEN votedate >= $yesterday THEN 1 ELSE 0 END) AS daycount,
SUM(CASE WHEN votedate >= $last_week THEN 1 ELSE 0 END) AS weekcount,
SUM(1) AS monthcount
FROM votes
WHERE yes_no = ‘YES’
AND votedate >= $last_month
GROUP BY id

Then make that a subquery and you can get the max values for the counts.

(Please allowing for the usual syntax sloppiness inherent in an untested query.)

---

In response to the comments:

To use it as an efficient subquery (i.e. not correlated) to get the maximum values:

SELECT
MAX(daycount) AS MaxDayCount,
MAX(weekcount) AS MaxWeekCount,
MAX(monthcount) AS MaxMonthCount
FROM
(
…. all that stuff …
) AS qcounts

but of course you can’t attribute them to ids, because they are different. If you want them one at a time with ids, you might

SELECT id, monthcount
FROM
(
…. all that stuff …
) AS qcounts
ORDER BY monthcount DESC
LIMIT 1

and do it three times, once for day/week/month.

Note: this is all to illustrate some things you could accomplish in a single reasonably efficient query. I wouldn’t be surprised if you were to find it’s simplest (and simple == good) to break it up as others suggest.