Question

我有一个包含这些文件的表：

**Date           Name       Table      Direction      Code**
13/01/1978     Jacks      xxxx       stret ...      1
13/01/1978     John       xxxx       xxxxx          0
...........
12/01/1978     Dave       xxxx       xxxxxx         0
12/01/1978     Suse       xxxxx      xxxxxx         0
...........
11/01/1978     Mickey     xxxx       xxxxx          1
11/01/1978     Suse       xxxxxx     xxxxxxx        2
11/01/1978     Nune       xxxxxx     xxxxxxx        2
.....
09/01/1978     .....      ......     .......        1
08/01/1978     .....      ......     .......        0
21/01/1978     .....      ......     .......        1
.....

我想提取按日期后代分组的三个第一个事件，并检测字段代码在具有相同日期的任何行中是否具有值“1”。我想要的结果是：

The first date is 21/01/1978, and have code 1
The second date is 13/01/1978 and have code 1
The third date is 12/01/1978 and don't have code 1

我正在尝试rownum但不行。请问你能帮帮我吗？谢谢，抱歉我的英文！

Answer 1

我不知道为什么rownum不能为你工作，也许因为你试图在订单之前插入内部查询（你首先要订购，然后才使用rownum因为每一行会得到另一个rownum）

SELECT Date,code from (
(SELECT Date,max(Code) as code from YourTable
GROUP BY Date
ORDER BY Date DESC
)
WHERE rownum <= 3

Answer 2

试试这希望它有帮助

SELECT DISTINCT B.DT,
  DECODE(B.VL,1,' has code 1',999,' dont have code 1') inf
FROM
  (SELECT A.DT,
    CASE
      WHEN A.CODE = 1
      THEN 1
      ELSE 999
    END VL,
    ROW_NUMBER() OVER(PARTITION BY A.DT ORDER BY A.CODE DESC) RN
  FROM
    (SELECT '01/03/2016' AS DT,1 AS CODE FROM DUAL
    UNION ALL
    SELECT '01/03/2016' AS DT,0 AS CODE FROM DUAL
    UNION ALL
    SELECT '05/03/2016' AS DT,0 AS CODE FROM DUAL
    UNION ALL
    SELECT '05/03/2016' AS DT,0 AS CODE FROM DUAL
    UNION ALL
    SELECT '06/03/2016' AS DT,1 AS CODE FROM DUAL
    )A
  ) B
  WHERE B.RN = 1;

Answer 3

你可以用很多方式做到这一点，其中一种就是：

select date, 
       decode(date_cnt, 0, 'dont have code = 1', 'have code = 1) 
from (
  select t.date, 
         max(t.code) as date_cnt
     from table_name t
     group by date
     order by date desc)
where rownum <=3

这样你首先在内部查询中按日期对它们进行排序，以及准备有关代码= 1的信息。外部查询将获取前3行并将代码更改为所需信息（如果0 =＆gt;没有代码，否则就有代码）

查询以获得最多3个结果

3 个答案: