的Oracle SQL - 獨特的價值屬於一年的總和，首次

Question

遺憾的冠軍，但有點難以解釋在連續事務所的話題..

我有一個這樣的表，我想知道（每年的每個月）第一次獲得獎金的員工人數。

EMPLOYEE_NAME MONTH BONUS_RECEIVED 
AAA     1    1 
BBB     1    1 

CCC     2    1 
AAA     2    1 
DDD     2    1 

AAA     3    1 
BBB     3    1 
XXX     3    1 

所以，結果應該是

MONTH  TOTAL_BONUS 
1      2 
2      2 
3      1 

• 月1，僱員AAA和BBB接收獎金（所以結果是2）

• 第2個月，僱員CCC和DD收到獎金（AAA已經收到全年），所以結果是2

• 第3個月，只有員工XXX h爲獲得獎金，因爲AAA和BBB已經收到它在全年的

Answer 1

您可以使用ROW_NUMBER（）解析函數。

例如，

設置

CREATE TABLE t 
    (EMPLOYEE_NAME varchar2(3), MONTH number, BONUS_RECEIVED number); 

INSERT ALL 
    INTO t (EMPLOYEE_NAME, MONTH, BONUS_RECEIVED) 
     VALUES ('AAA', 1, 1) 
    INTO t (EMPLOYEE_NAME, MONTH, BONUS_RECEIVED) 
     VALUES ('BBB', 1, 1) 
    INTO t (EMPLOYEE_NAME, MONTH, BONUS_RECEIVED) 
     VALUES ('CCC', 2, 1) 
    INTO t (EMPLOYEE_NAME, MONTH, BONUS_RECEIVED) 
     VALUES ('AAA', 2, 1) 
    INTO t (EMPLOYEE_NAME, MONTH, BONUS_RECEIVED) 
     VALUES ('DDD', 2, 1) 
    INTO t (EMPLOYEE_NAME, MONTH, BONUS_RECEIVED) 
     VALUES ('AAA', 3, 1) 
    INTO t (EMPLOYEE_NAME, MONTH, BONUS_RECEIVED) 
     VALUES ('BBB', 3, 1) 
    INTO t (EMPLOYEE_NAME, MONTH, BONUS_RECEIVED) 
     VALUES ('XXX', 3, 1) 
SELECT * FROM dual; 

查詢

SQL> SELECT MONTH, 
    2 COUNT(rn) total_bonus 
    3 FROM 
    4 (SELECT t.*, 
    5  row_number() OVER(PARTITION BY employee_name ORDER BY MONTH) rn 
    6 FROM t 
    7 WHERE BONUS_RECEIVED = 1 
    8 ) 
    9 WHERE rn = 1 
10 GROUP BY MONTH; 

    MONTH TOTAL_BONUS 
---------- ----------- 
     1   2 
     2   2 
     3   1 

Answer 2

雙聚合解決您的問題：

select month, count(1) as total_bonus 
from (
    select employee_name, min(month) as month 
    from table_like_this 
    where bonus_received = 1 
    group by employee_name 
) 
group by month; 

首先，爲每個員工找到他/她收到的第一個月獎金。然後，您計算每個「找到的第一個獎金收到月份」的員工數量。

Answer 3

U可以也使用RANK（）

SELECT MONTH 
     ,COUNT(BONUS) AS BONUS 
    FROM (
     SELECT EMPLOYEE 
      ,MONTH 
      ,BONUS 
      ,RANK() OVER (
       PARTITION BY EMPLOYEE ORDER BY MONTH 
       ) AS RN 
     FROM TBTEST 
     ) 
    WHERE RN = 1 
    GROUP BY MONTH