Question

我有四个这样的列。

Material  Description  Quantity         Date
    a         133       200     26-09-2016 12:33
    a         133       400     27-09-2016 10:33

我需要在Max（Date）;

时从该材料中获取数量

我试过这个但是如果数量不同则显示两行。

Select material , description , quantity , max(date)
FROM  materials
group by material, description , quantity

Answer 1

在WHERE子句中使用该条件，如

Select material , description , quantity 
FROM  materials
WHERE "Date"  = (select max("Date") from materials)

Answer 2

使用RANK() analytic function：

SELECT *
FROM   (
  SELECT materials,
         description,
         quantity,
         date,
         RANK() OVER ( PARTITION BY materials ORDER BY date DESC ) AS rnk 
  FROM   materials
)
WHERE  rnk = 1;

如果存在具有相同materials和最大date值的行，则会获得多行 - 如果您只需要一行，则使用ROW_NUMBER()代替RANK()

Answer 3

你可以像这样使用row_number()（如果你需要找到组中所有材料的数量＆＃34; a＆＃34;＆＃34; b＆＃34; ）：

WITH a(Material, Description , Quantity , sDate) AS
  (SELECT 'b',    133,    1200 ,    to_date('26-09-2016 12:33','dd-mm-yyyy hh24:mi')  FROM dual  UNION ALL
   SELECT 'b',    133,    2200 ,    to_date('29-09-2016 12:33','dd-mm-yyyy hh24:mi')  FROM dual  UNION ALL
   SELECT 'a',    133,    200 ,    to_date('26-09-2016 12:33','dd-mm-yyyy hh24:mi')  FROM dual  UNION ALL
   SELECT 'a',    133,    400 ,    to_date('27-09-2016 10:33','dd-mm-yyyy hh24:mi')  FROM dual  )
SELECT *
  FROM (SELECT a.*,
               row_number() over(partition BY material order by sdate DESC) rn
          FROM a)
 WHERE rn = 1


MATERIAL DESCRIPTION   QUANTITY SDATE             RN
-------- ----------- ---------- --------- ----------
a                133        400 27-SEP-16          1 
b                133       2200 29-SEP-16          1

Answer 4

SELECT *
FROM   (
  SELECT materials,
         description,
         quantity,
         date,
         RANK() OVER ( PARTITION BY materials ORDER BY date DESC ) AS rnk 
  FROM   materials
)
WHERE  rnk = 1;

当另一列在oracle中具有不同的值时，如何选择MAX（列）行？

4 个答案: