sql - 从每个组中选择第一行，其中 ORDER BY 多于一列

Question

我有一个T带有列x的表y，，，，a这样b

SELECT x,y,a,b FROM T
ORDER BY x,y,a,b

给我下表

 x    | y    | a    | b
 x1   | y1   | a1   | b1
 x1   | y1   | a1   | b2
 x1   | y1   | a2   | b1
 x1   | y2   | a1   | b1
 x1   | y2   | a1   | b2
 x1   | y2   | a2   | b1

我将如何获得每个 x,y 组的第一行？也就是说，我将如何获得下表

 x    | y    | a    | b
 x1   | y1   | a1   | b1
 x1   | y2   | a1   | b1

---

这是第二个例子：对于一个表 T 使得

 x    | y    | a    | b
 x1   | y1   | a1   | b3
 x1   | y1   | a1   | b4
 x1   | y1   | a2   | b1
 x1   | y1   | a2   | b2
 x1   | y2   | a1   | b3
 x1   | y2   | a1   | b4
 x1   | y2   | a2   | b1
 x1   | y2   | a2   | b2

我期待得到

 x    | y    | a    | b
 x1   | y1   | a1   | b3
 x1   | y2   | a1   | b3

Answer 1

考虑 SQL 2005 或更高版本：

SELECT T1.X,
       T1.Y,
       T1.A,
       T1.B
  FROM
      (SELECT X,
              Y,
              A,
              B,
              ROW_NUMBER() OVER (Partition BY X,Y Order By A,B) AS RowNum
         FROM T
     ) T1
WHERE T1.RowNum = 1

Answer 2

根据您的评论和新表结构，您可以使用这个与 DBMS 无关的简单解决方案：

SELECT a.x,a.y,a.a,MIN(a.b) b
FROM T a
JOIN ( SELECT x,y,MIN(a) a FROM T GROUP BY x,y ) b ON
    a.x = b.x AND
    a.y = b.y AND
    a.a = b.a
GROUP BY a.x,a.y,a.a

---

SQL-Fiddle 演示

Answer 3

这是一个非常有效的Oracle解决方案

select x, 
       y, 
       min(a) as a, 
       min(b) keep( dense_rank first order by a) as b
  from T
 group by x,y

Answer 4

请试试这个，让我知道

select x,y,MIN(a) a,MIN(b) b
from T
group by X,y