Top 20 percent by id - MySQL - mysql

Question

I am using a modified version of a query similiar to another question here:Convert SQL Server query to MySQL
Select *
from
(
SELECT tbl.*, #counter := #counter +1 counter
FROM (select #counter:=0) initvar, tbl
Where client_id = 55
ORDER BY ordcolumn
) X
where counter >= (80/100 * #counter);
ORDER BY ordcolumn
tbl.* contains the field 'client_id' and I am attempting to get the top 20% of the records for each client_id in a single statement. Right now if I feed it a single client_id in the where statement it gives me the correct results, however if I feed it multiple client_id's it simply takes the top 20% of the combined recordset instead of doing each client_id individually.
I'm aware of how to do this in most databases, but the logic in MySQL is eluding me. I get the feeling it involves some ranking and partitioning.
Sample data is pretty straight forward.
Client_id rate
1 1
1 2
1 3
(etc to rate = 100)
2 1
2 2
2 3
(etc to rate = 100)
Actual values aren't that clean, but it works.
As an added bonus...there is also a date field associated to these records and 1 to 100 exists for this client for multiple dates. I need to grab the top 20% of records for each client_id, year(date),month(date)

Answer 1

You need to do the enumeration for each client:
SELECT *
FROM (SELECT tbl.*, #counter := #counter +1 counter
(#rn := if(#c = client_id, #rn + 1,
if(#c := client_id, 1, 1)
)
)
FROM (select #c := -1, #rn := 0) initvar CROSS JOIN tbl
ORDER BY client_id, ordcolumn
) t cross join
(SELECT client_id, COUNT(*) as cnt
FROM tbl
GROUP BY client_id
) tt
where rn >= (80/100 * tt.cnt);
ORDER BY ordcolumn;

Answer 2

Using Gordon's answer as a starting point, I think this might be closer to what you need.
SELECT t.*
, (#counter := #counter+1) AS overallRow
, (#clientRow := if(#prevClient = t.client_id, #clientRow + 1,
if(#prevClient := t.client_id, 1, 1) -- This just updates #prevClient without creating an extra field, though it makes it a little harder to read
)
) AS clientRow
-- Alteratively (for everything done in clientRow)
, #clientRow := if(#prevClient = t.client_id, #clientRow + 1, 1) AS clientRow
, #prevClient := t.client_id AS extraField
-- This may be more reliable as well; I not sure if the order
-- of evaluation of IF(,,) is reliable enough to guarantee
-- no side effects in the non-"alternatively" clientRow calculation.
FROM tbl AS t
INNER JOIN (
SELECT client_id, COUNT(*) AS c
FROM tbl
GROUP BY client_id
) AS cc ON tbl.client_id = cc.client_id
INNER JOIN (select #prevClient := -1, #clientRow := 0) AS initvar ON 1 = 1
WHERE t.client_id = 55
HAVING clientRow * 5 < cc.c -- You can use a HAVING without a GROUP BY in MySQL
-- (note that clientRow is derived, so you cannot use it in the `WHERE`)
ORDER BY t.client_id, t.ordcolumn
;