仅用 NULL 替换连续记录中的重复值 [重复]

Question

假设我有两张桌子
Tran (Tran_Id, Tran_Name) TFlag(Tran_Id, Flag)

我的查询结果是这样的：

TRAN_ID TRAN_NAME   FLAG    
-------------------------
101     Lend        A   
101     Lend        B   
101     Lend        C   
101     Lend        D   
102     Borrow      E   
101     Lend        F   
101     Lend        G   

我希望输出是这样的：

TRAN_ID TRAN_NAME   FLAG
-------------------------
101     Lend        A   
(null)  (null)      B   
(null)  (null)      C   
(null)  (null)      D   
102     Borrow      E   
101     Lend        F   
(null)  (null)      G   

Answer 1

您可以使用分析 LAG() OVER()。

SQL Fiddle 用于工作演示。

例如，

SQL> SELECT
  2    CASE
  3      WHEN lag(tran_id) over(order by NULL) = tran_id
  4      THEN NULL
  5      ELSE tran_id
  6    END tran_id,
  7    CASE
  8      WHEN lag(tran_name) over(order by NULL) = tran_name
  9      THEN NULL
 10      ELSE tran_name
 11    END tran_name,
 12    flag
 13  FROM t;

   TRAN_ID TRAN_N F
---------- ------ -
       101 Lend   A
                  B
                  C
                  D
       102 Borrow E
       101 Lend   F
                  G

7 rows selected.

Answer 2

在 SQL Server 和 Oracle 中都可以使用 LAG 分析函数。甲骨文：

WITH data (tran_id, tran_name, flag) AS (
    SELECT  01,   'Lend', 'A' FROM DUAL UNION ALL
    SELECT 101,   'Lend', 'B' FROM DUAL UNION ALL
    SELECT 101,   'Lend', 'C' FROM DUAL UNION ALL
    SELECT 101,   'Lend', 'D' FROM DUAL UNION ALL
    SELECT 102, 'Borrow', 'E' FROM DUAL UNION ALL
    SELECT 101,   'Lend', 'F' FROM DUAL UNION ALL
    SELECT 101,   'Lend', 'G' FROM DUAL 
)
SELECT
    NVL2(keep, tran_id, null) tran_id,
    NVL2(keep, tran_name, null) tran_name,
    flag
FROM (
    SELECT
        tran_id, tran_name, flag,
        CASE WHEN LAG(tran_id) OVER (ORDER BY flag) <> tran_id OR LAG(tran_name) OVER (ORDER BY flag) <> tran_name THEN 1 END keep
    FROM
        data)