Count records which fulfil given condition - mysql

Question

I have a table with the following schema:
+-------------------------------------------------------+
| table_counter |
+----+---------------------+------------+---------------+
| id | timestamp | entry_type | country |
+----+---------------------+------------+---------------+
+----+---------------------+------------+---------------+
| 10 | 2017-05-01 12:00:00 | click | Germany |
+----+---------------------+------------+---------------+
| 11 | 2017-05-01 12:00:00 | view | Austria |
+----+---------------------+------------+---------------+
| 12 | 2017-05-01 12:00:00 | click | UK |
+----+---------------------+------------+---------------+
| 13 | 2017-05-01 12:00:00 | view | USA |
+----+---------------------+------------+---------------+
I need to return the following result: Select the sum of views and clicks of the top 5 countries by sum of views in the past 30 days.
I know how to count the records all right, but how do I define the constrains? How do I return all entries from five countries with the highest number of views?
Limiting the result to the last 30 days is trivial, but I'm pretty much stuck at the beginning.

Answer 1

Using order by and limit keywords,
SELECT SUM(IF(entry_type = "view", 1, 0)) as view_count FROM t3 GROUP BY country, entry_type ORDER BY view_count DESC LIMIT 5
--EDIT
As per the requirement stated in the comments, here's the updated query:
SELECT SUM(view_click_count) as all_total FROM (SELECT country, SUM(IF(entry_type = "view", 1, 0)) as view_count, SUM(IF(entry_type = "click", 1, 0)) as click_count, count(entry_type) as view_click_count FROM t3 GROUP BY country ORDER BY view_count DESC LIMIT 5) t2
all_total gives the total count as needed, for top 5 countries.

Answer 2

You can do it this way:
select
tc.country,
count(case entry_type when 'click' then 1 else null end) clicks,
count(case entry_type when 'view' then 1 else null end) views
from table_counter tc
inner join (
select top 5 country from [dbo].[table_counter]
where entry_type = 'view'
and timestamp >= DATEADD(DAY, -30, GETDATE())
group by country
order by count(entry_type) desc
) t on t.country = tc.country
where timestamp >= DATEADD(DAY, -30, GETDATE())
group by tc.country
order by views desc
This is for SQL Server. A few tweaks might be needed for MySQL (i.e. 'Limit' instead of 'TOP')

Answer 3

You can get top 5 countries by views with the following query, e.g.:
SELECT country, count(*) as 'views'
FROM table
WHERE timestamp BETWEEN DATE_SUB(NOW(), INTERVAL 1 MONTH) AND NOW()
AND entry_type = 'view'
GROUP BY country
ORDER BY count(*) DESC
LIMIT 5
Now, to select clicks, you can add another query in SELECT , e.g.:
SELECT t.country, COUNT(*) as 'views',
(SELECT COUNT(*)
FROM `table`
WHERE country = t.country
AND entry_type = 'click'
AND timestamp BETWEEN DATE_SUB(NOW(), INTERVAL 1 MONTH) AND NOW()
) as 'clicks'
FROM `table` t
WHERE t.timestamp BETWEEN DATE_SUB(NOW(), INTERVAL 1 MONTH) AND NOW()
AND t.entry_type = 'view'
GROUP BY t.country
ORDER BY count(*) DESC
LIMIT 5
Here's the SQL Fiddle.
Update
To get the SUM of views and clicks, wrap the above query into another SELECT, e.g.:
SELECT country, views + clicks
FROM(
SELECT t.country, COUNT(*) as 'views',
(SELECT COUNT(*)
FROM `table`
WHERE country = t.country
AND entry_type = 'click'
AND timestamp BETWEEN DATE_SUB(NOW(), INTERVAL 1 MONTH) AND NOW()
) as 'clicks'
FROM `table` t
WHERE t.timestamp BETWEEN DATE_SUB(NOW(), INTERVAL 1 MONTH) AND NOW()
AND t.entry_type = 'view'
GROUP BY t.country
ORDER BY count(*) DESC
LIMIT 5
) b;
Here's the updated SQL Fiddle.