MySQL-如何获取特定“ id”内一列中所有行的差值？

Question

For example, in this table:

n   id   num

1   10   100
2   11    60
3   10    20
4   10    20
5   11    10

How do I subtract all the values of num (from top to bottom) with id = 10?

The first value of num in id=10 is subtracted to the 2nd value of num in id = 10 and the answer is subtracted to the third value of num in id = 10 (an so on and so forth if there are n numbers of num with the id =10)

It should display this:

difference of id = 10

50

Answer 1

Here is the working code you need:

SELECT AA.ID, (num_duo-total_sum) as num_diff FROM 

(SELECT id, 2*num as num_duo, MIN(n) FROM t1 GROUP BY id ) AA LEFT JOIN 


(SELECT id, SUM(num) as total_sum FROM t1 GROUP BY id) BB ON BB.id = AA.id 

SQL FIDDLE

Answer 2

If I understand correctly, you want to take the first value for the given id and then subtract subsequent values. "first" is defined by the first column, n .

If so, then this might help:

select sum(case when t.n = f.firstn then num else - num end)
from table t cross join
     (select min(n) as firstn
      from table t
      where id = 10
     ) f
where id = 10;

Here is a SQL Fiddle.