部分和

Question

我按照表：

2016-01  111.00  100.00 
2016-01  111.00  11.00 
2016-02  222.00  100.00 
2016-02  222.00  100.00 
2016-02  222.00  22.00 
2016-01  333.00  200.00 
2016-01  333.00  133.00 

我需要：

2016-01  444.00 
2016-02  222.00 

我怎麼能接受這個數據？按期間求和，但只有明顯的價值？

Answer 1

使用SUM()與DISTINCT

SELECT DateColName ,SUM(DISTINCT ColName) 
FROM TABLENAME 
GROUP BY DateColName 

Answer 2

;with cte 
as 
(select 
date,col1,col2 
,row_number() over (partition by date,col1 order by date) as rownum 
) 
select 
date,sum(col1) 
from cte 
where rownum=1 
group by date 

Answer 3

SUM() + DISTINCT + GROUP BY是你的朋友。

重新創建測試用例：

sqlite> CREATE TABLE money(month TEXT, q1 FLOAT, q2 FLOAT); 
sqlite> INSERT INTO money VALUES('2016-01','111.00','100.00'); 
sqlite> INSERT INTO money VALUES('2016-01','111.00','11.00'); 
sqlite> INSERT INTO money VALUES('2016-02','222.00','100.00'); 
sqlite> INSERT INTO money VALUES('2016-02','222.00','100.00'); 
sqlite> INSERT INTO money VALUES('2016-02','222.00','22.00'); 
sqlite> INSERT INTO money VALUES('2016-01','333.00','200.00'); 
sqlite> INSERT INTO money VALUES('2016-01','333.00','133.00'); 

檢查數據是一樣的你：

sqlite> SELECT * FROM money; 
2016-01|111.0|100.0 
2016-01|111.0|11.0 
2016-02|222.0|100.0 
2016-02|222.0|100.0 
2016-02|222.0|22.0 
2016-01|333.0|200.0 
2016-01|333.0|133.0 

運行查詢：

sqlite> SELECT month, SUM(DISTINCT q1) FROM money GROUP BY month; 
2016-01|444.0 
2016-02|222.0