總和加入重複行（SQL）時

Question

我有查詢結果如下：

SELECT ... ON CIA_factbook_dataset.my_name = World_Bank_dataset.my_name ... 

+----------------+------+-------------+-----------------+---------+--------+ 
| my_name  | Year | CIA_name | World_Bank_name | CIA_GDP | WB_GDP | 
+----------------+------+-------------+-----------------+---------+--------+ 
| United Kingdom | 2010 | UK   | United Kingdom | 2850 | 2800 | 
| United Kingdom | 2010 | UK   | Channel Islands | 2850 | 11 | 
| Cyprus   | 2010 | CYPRUS TURK | CYPRUS TURK  | 22 | 22 | 
| Cyprus   | 2010 | CYPRUS TURK | CYPRUS GRK  | 22 | 33 | 
| Cyprus   | 2010 | CYPRUS GRK | CYPRUS TURK  | 33 | 22 | 
| Cyprus   | 2010 | CYPRUS GRK | CYPRUS GRK  | 33 | 33 | 
+----------------+------+-------------+-----------------+---------+--------+ 

我需要計算的分國別數據的總和，但如果我只用GROUP BY my_name,year，它計算幾次相同數目的總和。

最終的結果應該是：

+----------------+------+---------+--------+ 
| my_name  | Year | CIA_GDP | WB_GDP | 
+----------------+------+---------+--------+ 
| United Kingdom | 2010 | 2850 | 2811 | 
| Cyprus   | 2010 | 55 | 55 | 
+----------------+------+---------+--------+ 

相反的：

+----------------+------+---------+--------+ 
| my_name  | Year | CIA_GDP | WB_GDP | 
+----------------+------+---------+--------+ 
| United Kingdom | 2010 | 5700 | 2811 | 
| Cyprus   | 2010 | 110 | 110 | 
+----------------+------+---------+--------+ 

如何實現這一目標？
比使用SUM(distinct CIA_GDP),SUM(distinct WB_GDP)更好的方法嗎？
（理論上，土耳其塞浦路斯和希臘塞浦路斯的GDP可能是相同的）

Answer 1

對於這個我假設my_name，Year是兩個表中是唯一的。

SQL Fiddle

SELECT t1.my_name, t1.YEAR, SUM_CIA_GDP, SUM_WB_GDP 
FROM (
    SELECT DISTINCT my_name, YEAR, SUM(CIA_GDP) AS SUM_CIA_GDP 
    FROM t 
    GROUP BY my_name, YEAR, WB_GDP 
    ) t1 
JOIN ( 
    SELECT DISTINCT my_name, YEAR, SUM(WB_GDP) AS SUM_WB_GDP 
    FROM t 
    GROUP BY my_name, YEAR, CIA_GDP 
    ) t2 
    ON t1.my_name = t2.my_name 
     AND t1.YEAR = t2.YEAR 

Results：

|  my_name | YEAR | SUM_CIA_GDP | SUM_WB_GDP | 
|----------------|------|-------------|------------| 
|   Cyprus | 2010 |   55 |   55 | 
| United Kingdom | 2010 |  2850 |  2811 | 

Answer 2

SQL Fiddle

的MySQL 5.6架構設置：

CREATE TABLE t 
    (`my_name` varchar(14), `Year` int, `CIA_name` varchar(11), `World_Bank_name` varchar(15), `CIA_GDP` int, `WB_GDP` int) 
; 

INSERT INTO t 
    (`my_name`, `Year`, `CIA_name`, `World_Bank_name`, `CIA_GDP`, `WB_GDP`) 
VALUES 
    ('United Kingdom', 2010, 'UK', 'United Kingdom', 2850, 2800), 
    ('United Kingdom', 2010, 'UK', 'Channel Islands', 2850, 11), 
    ('Cyprus', 2010, 'CYPRUS TURK', 'CYPRUS TURK', 22, 22), 
    ('Cyprus', 2010, 'CYPRUS TURK', 'CYPRUS GRK', 22, 33), 
    ('Cyprus', 2010, 'CYPRUS GRK', 'CYPRUS TURK', 33, 22), 
    ('Cyprus', 2010, 'CYPRUS GRK', 'CYPRUS GRK', 33, 33) 
; 

查詢1：

SELECT my_name, Year, SUM(CIA_GDP), WB_GDP 
FROM (
    SELECT my_name, Year, CIA_GDP, SUM(WB_GDP) WB_GDP 
    FROM t 
    GROUP BY my_name, Year, CIA_GDP 
) t1 
GROUP BY my_name, Year, WB_GDP 

Results：

|  my_name | Year | SUM(CIA_GDP) | WB_GDP | 
|----------------|------|--------------|--------| 
|   Cyprus | 2010 |   55 |  55 | 
| United Kingdom | 2010 |   2850 | 2811 |