Question

我有一个审计表，将用户状态保持为更改时的状态。表格如下所示。

{z: 'data'}

每个用户都可能无法转换到应用程序中的每个状态，即可。我想要做的是捕获每个用户更改时的日期和状态，并将其放在一行中。

所以输出我期待它如下所示。

{pitch1: 'data'}

我一直在阅读有关pivot和unpivot以及解码但我不确定这是否在oracle中是可行的。如果是的话，有人可以引导我走上正确的道路吗？

Answer 1

以下内容可能更容易理解：

WITH cteID AS (SELECT DISTINCT USER_ENTL_ID AS ID FROM AUDIT_TABLE),
     cteS  AS (SELECT USER_ENTL_ID AS ID,
                      'S' AS S_STATUS,
                      MIN(CREATED_DATE) AS S_CREATED
                 FROM AUDIT_TABLE
                 WHERE STATUS = 'S'
                 GROUP BY USER_ENTL_ID),
     cteC AS  (SELECT USER_ENTL_ID AS ID,
                      'C' AS C_STATUS,
                      MIN(CREATED_DATE) AS C_CREATED
                 FROM AUDIT_TABLE
                 WHERE STATUS = 'C'
                 GROUP BY USER_ENTL_ID),
     cteSP AS (SELECT USER_ENTL_ID AS ID,
                      'SP' AS SP_STATUS,
                      MIN(CREATED_DATE) AS SP_CREATED
                 FROM AUDIT_TABLE
                 WHERE STATUS = 'SP'
                 GROUP BY USER_ENTL_ID),
     cteW AS  (SELECT USER_ENTL_ID AS ID,
                      'W' AS W_STATUS,
                      MIN(CREATED_DATE) AS W_CREATED
                 FROM AUDIT_TABLE
                 WHERE STATUS = 'W'
                 GROUP BY USER_ENTL_ID)
SELECT i.ID,
       s.S_STATUS,
       s.S_CREATED,
       c.C_STATUS,
       c.C_CREATED,
       sp.SP_STATUS,
       sp.SP_CREATED,
       w.W_STATUS,
       w.W_CREATED
  FROM cteID i
  LEFT OUTER JOIN cteS s
    ON s.ID = i.ID
  LEFT OUTER JOIN cteC c
    ON c.ID = i.ID
  LEFT OUTER JOIN cteSP sp
    ON sp.ID = i.ID
  LEFT OUTER JOIN cteW w
    ON w.ID = i.ID

祝你好运。

Answer 2

使用＆＃34;条件聚合＆＃34;是处理此类需求的传统且有效的方式：

SQL Fiddle

PostgreSQL 9.6架构设置：

CREATE TABLE AUDIT_TABLE
    (USER_ENTL_ID int, USER_STATUS varchar(2), CREATED_DATE timestamp)
;

INSERT INTO AUDIT_TABLE
    (USER_ENTL_ID, USER_STATUS, CREATED_DATE)
VALUES
    (1, 'S', '2017-10-20 00:00:00'),
    (1, 'C', '2017-10-21 00:00:00'),
    (1, 'W', '2017-10-22 00:00:00'),
    (1, 'SP', '2017-10-23 00:00:00'),
    (2, 'S', '2017-10-24 00:00:00'),
    (2, 'C', '2017-10-25 00:00:00')
;

查询1 ：

nb，根据您的数据使用MIN或MAX可能很重要，但是每个输出位置的数据中只有一个值，那么可以使用任一函数。

SELECT
      USER_ENTL_ID
    , MAX(CASE WHEN USER_STATUS = 'S' THEN USER_STATUS END) s_status
    , MIN(CASE WHEN USER_STATUS = 'S' THEN CREATED_DATE END) s_created
    , MAX(CASE WHEN USER_STATUS = 'C' THEN USER_STATUS END) c_status
    , MIN(CASE WHEN USER_STATUS = 'C' THEN CREATED_DATE END) c_created
    , MAX(CASE WHEN USER_STATUS = 'W' THEN USER_STATUS END) w_status
    , MIN(CASE WHEN USER_STATUS = 'W' THEN CREATED_DATE END) w_created
    , MAX(CASE WHEN USER_STATUS = 'SP' THEN USER_STATUS END) sp_status
    , MIN(CASE WHEN USER_STATUS = 'SP' THEN CREATED_DATE END) sp_created
FROM AUDIT_TABLE
GROUP BY 
      USER_ENTL_ID

<强> Results ：

| user_entl_id | s_status |            s_created | c_status |            c_created | w_status |            w_created | sp_status |           sp_created |
|--------------|----------|----------------------|----------|----------------------|----------|----------------------|-----------|----------------------|
|            1 |        S | 2017-10-20T00:00:00Z |        C | 2017-10-21T00:00:00Z |        W | 2017-10-22T00:00:00Z |        SP | 2017-10-23T00:00:00Z |
|            2 |        S | 2017-10-24T00:00:00Z |        C | 2017-10-25T00:00:00Z |   (null) |               (null) |    (null) |               (null) |

ADDED

进一步解释：如果您删除了MIN或MAX功能并删除了分组，这就是您所得到的：

+--------------+----------+----------------------+----------+----------------------+----------+----------------------+-----------+----------------------+
| user_entl_id | s_status |      s_created       | c_status |      c_created       | w_status |      w_created       | sp_status |      sp_created      |
+--------------+----------+----------------------+----------+----------------------+----------+----------------------+-----------+----------------------+
|            1 | S        | 2017-10-20T00:00:00Z | (null)   | (null)               | (null)   | (null)               | (null)    | (null)               |
|            1 | (null)   | (null)               | C        | 2017-10-21T00:00:00Z | (null)   | (null)               | (null)    | (null)               |
|            1 | (null)   | (null)               | (null)   | (null)               | W        | 2017-10-22T00:00:00Z | (null)    | (null)               |
|            1 | (null)   | (null)               | (null)   | (null)               | (null)   | (null)               | SP        | 2017-10-23T00:00:00Z |
|            2 | S        | 2017-10-24T00:00:00Z | (null)   | (null)               | (null)   | (null)               | (null)    | (null)               |
|            2 | (null)   | (null)               | C        | 2017-10-25T00:00:00Z | (null)   | (null)               | (null)    | (null)               |
+--------------+----------+----------------------+----------+----------------------+----------+----------------------+-----------+----------------------+

如果您研究过，您会看到我们关心的数据每行只有一个值（，每个USER_ENTL_ID ），但它们分布在多行上。所以MIN / MAX功能和GROUP BY＆＃34;平坦＆＃34;结果，所以我们最终得到想要的结果。 QED

Answer 3

只需使用PIVOT：

SQL Fiddle

Oracle 11g R2架构设置：

CREATE TABLE AUDIT_TABLE (USER_ENTL_ID, USER_STATUS, CREATED_DATE) AS
SELECT 1, 'S',  TIMESTAMP '2017-10-20 00:00:00' FROM DUAL UNION ALL
SELECT 1, 'C',  TIMESTAMP '2017-10-21 00:00:00' FROM DUAL UNION ALL
SELECT 1, 'W',  TIMESTAMP '2017-10-22 00:00:00' FROM DUAL UNION ALL
SELECT 1, 'SP', TIMESTAMP '2017-10-23 00:00:00' FROM DUAL UNION ALL
SELECT 2, 'S',  TIMESTAMP '2017-10-24 00:00:00' FROM DUAL UNION ALL
SELECT 2, 'C',  TIMESTAMP '2017-10-25 00:00:00' FROM DUAL

查询1 ：

SELECT *
FROM   AUDIT_TABLE
PIVOT (
  MAX( Created_Date ) AS Created,
  MAX( User_Status ) AS  Status
  FOR User_Status IN (
    'S' AS S, 'C' AS C, 'W' AS W, 'SP' AS SP
  )
)

<强> Results ：

| USER_ENTL_ID |             S_CREATED | S_STATUS |             C_CREATED | C_STATUS |             W_CREATED | W_STATUS |            SP_CREATED | SP_STATUS |
|--------------|-----------------------|----------|-----------------------|----------|-----------------------|----------|-----------------------|-----------|
|            1 | 2017-10-20 00:00:00.0 |        S | 2017-10-21 00:00:00.0 |        C | 2017-10-22 00:00:00.0 |        W | 2017-10-23 00:00:00.0 |        SP |
|            2 | 2017-10-24 00:00:00.0 |        S | 2017-10-25 00:00:00.0 |        C |                (null) |   (null) |                (null) |    (null) |

将oracle表行展平为具有多列的单行

3 个答案:

ADDED