MySQL subtract two count columns - mysql

Question

I have a table like this:
client msg_type msg_body id
------ -------- -------- ---
123 typeA success abc
123 typeB success abc
456 typeA success abc
456 typeB failure abc
123 typeA success abc
123 typeA success abc
789 typeA success def
789 typeB success def
etc.
I would like output like this:
client diff id
------ ---- ---
123 2 abc
456 1 abc
789 0 def
where diff is the count of typeA:success messages - typeB:success messages. I can get the count of the typeA success using something like:
select client, count(*) from mytable
where msg_type="typeA" and msg_body="success"
However, I can't figure out how to put another count in there (for typeB) and also subtract.
I tried something like:
select client, count(*) from mytable
where msg_type="typeA" and msg_body="success" - count(*)
from mytable where msg_type="typeB" and msg_body="success"
But of course it didn't work, or I wouldn't be asking here. :) Any advice?
Edit: added another column. I tried the two suggestions given, but it only seems to return the results for one of the ids, not both.
Edit #2: I tried wrapping the SELECT query with:
select id, count(*) from (select ...) as anothertable where count_a_minus_count_b = 0;
I was hoping the output would be like:
id count
--- -----
abc 2
def 1
where count is the number of clients where the difference between typeA:success and typeB:success is 0.

Answer 1

COUNT counts non-null values, so you can construct an expression that's non-null when msg_type = 'typeA', and an expression that's non-null when msg_type = 'typeB'. For example:
SELECT client,
COUNT(CASE WHEN msg_type = 'typeA' THEN 1 END) AS count_a,
COUNT(CASE WHEN msg_type = 'typeB' THEN 1 END) AS count_b,
COUNT(CASE WHEN msg_type = 'typeA' THEN 1 END)
- COUNT(CASE WHEN msg_type = 'typeB' THEN 1 END) AS count_a_minus_count_b
FROM mytable
WHERE msg_body = 'success'
GROUP
BY client
;
(Disclaimer: not tested.)

Answer 2

Another way:
SELECT
d.client, COALESCE(a.cnt, 0) - COALESCE(b.cnt, 0) AS diff, d.id
FROM
( SELECT DISTINCT client, id
FROM mytable
) AS d
LEFT JOIN
( SELECT client, COUNT(*) AS cnt, id
FROM mytable
WHERE msg_type = 'typeA'
AND msg_body = 'success'
GROUP BY client, id
) AS a
ON d.client = a.client
AND d.id = a.id
LEFT JOIN
( SELECT client, COUNT(*) AS cnt, id
FROM mytable
WHERE msg_type = 'typeB'
AND msg_body = 'success'
GROUP BY client, id
) AS b
ON d.client = b.client
AND d.id = b.id ;
Tested at SQL-Fiddle

Answer 3

Here you go:
select client,
(sum(case when msg_type='typeA' and msg_body='success' then 1 else 0 end) -
sum(case when msg_type='typeB' and msg_body='success' then 1 else 0 end)) as diff
from your_table
group by client

Answer 4

Here's one way to get the result:
SELECT t.client
, SUM(t.msg_type<=>'typeA' AND t.msg_body<=>'success')
- SUM(t.msg_type<=>'typeB' AND t.msg_body<=>'success') AS diff
FROM mytable t
GROUP BY t.client
(The expressions in this query are MySQL specific; for a more portable query, use a less concise CASE expression to obtain an equivalent result.)
As more terse and obfuscated alternative to return the same result:
SELECT t.client
, SUM((t.msg_body<=>'success')*((t.msg_type<=>'typeA')+(t.msg_type<=>'typeB')*-1)) AS diff
FROM mytable t
GROUP BY t.client