Question

我有一张桌子：

items status count date
----- ------ ----- ----------
apple good   100   01/02/2017
apple good   200   03/02/2017
apple bad    50    02/02/2017
pear  good   100   04/02/2017

结果，我希望根据状态汇总计数并显示每个项目的最新日期：

items  count  date
-----  -----  ----------
apple  250    03/02/2017
pear   100    04/02/2017

Answer 1

试试这个：

SELECT items, SUM(count) count, MAX(date) date
FROM table
GROUP BY items

如果你想为每个状态分别求和，那么：

SELECT items, status, SUM(count) count, MAX(date) date
FROM table
GROUP BY items, status

Answer 2

使用sum，max和group by以及适用于总和正确好坏的情况

select items, sum(case when status = 'good' then count else -count end), max(date)
from my_table 
group by itmes

或者如果你不需要基于好的和坏的相对金额

select items, sum( count ), max(date)
from my_table 
group by itmes

Answer 3

我想你只是想：

select items, sum(count), max(date)
from t
group by items;

SQL条件聚合

3 个答案: