获取同一列的总值多行和总共两个不同的行（mysql查询）

Question

I have a table "records" [Records][1]

select typeofservice, 
       Coalesce(sum(incomecost,expensecost),0) 'incomecost,expensecost' 
from   records 
where  dateoftrans = '2016-11-22' 
group by typeofservice with rollup;

I want to be able to have the result on the second image and this is my query... and all of this transactions happened on one day. Please help

https://i.stack.imgur.com/fLYav.png

Answer 1

The following query assumes that, for a given type of service, you want to report an income when the sum of the income cost exceeds the sum of the expense cost, and vice-versa. In each case, the difference between income and expense is reported in the appropriate column, with NULL being reported in the alternate column.

SELECT typeofservice,
       CASE WHEN SUM(incomecost) - SUM(expensecost) > 0
            THEN SUM(incomecost) - SUM(expensecost)
            ELSE NULL END AS incomecost,
       CASE WHEN SUM(expensecost) - SUM(incomecost) > 0
            THEN SUM(expensecost) - SUM(incomecost)
            ELSE NULL END AS expensecost
FROM records
WHERE dateoftrans = '2016-11-22'
GROUP BY typeofservice
WITH ROLLUP

Answer 2

If only for second image's result, you can try sum(incomecost) and sum(expensecost) like this:

SELECT
    typeofservice,
    Coalesce(SUM(incomecost), 0) AS incomecost,
    Coalesce(SUM(expensecost), 0) AS expensecost
FROM records
GROUP BY typeofservice
WITH ROLLUP