Question

我有一个表格，可以保持 process_id 对应的状态和 created_timestamp 。一旦过程状态改变，就插入一行。因此，存在与 process_id 相同的行，因为有与之关联的状态。我想使用此数据创建另一个表格/视图，该数据的一行对应 process_id ，其当前状态和之前的状态。我需要为此创建一个Informatica作业，但SQL查询将同样有用。
样本输入：

Process_id | Status     | Created 
         1 | In_queue   | 2014-08-01 00:01:01 
         1 | Started    | 2014-08-01 01:03:01 
         1 | In_process | 2014-08-01 01:50:20 
         1 | Complete   | 2014-08-01 03:10:20 

Sample Output: 
Process_id | Previous_status | Current_status | Updated 
         1 | In_process      | Complete       | 2014-08-01 03:10:20

Answer 1

SELECT Process_id, Previous_status, Current_status, Updated 
FROM (
    SELECT
    Process_id, 
    Status AS Current_status, 
    Created AS Updated,
    @prev_state AS Previous_status,
    @prev_state := Status 
    FROM
    your_table t
    , (select @prev_state := null) var_init
    WHERE Process_id = 1
    ORDER BY Created
) sq

更新：

要为所有Process_id执行此操作，只需获取每个Process_id的最新记录，您可以使用它：

SELECT sq.Process_id, sq.Previous_status, sq.Current_status, sq.Updated 
FROM (
    SELECT
    Process_id, 
    Status AS Current_status, 
    Created AS Updated,
    @prev_state := if(@prev_process != Process_id, null, @prev_state),
    @prev_state AS Previous_status,
    @prev_state := Status,
    @prev_process := Process_id
    FROM
    your_table t
    , (select @prev_state := null, @prev_process := null) var_init
    ORDER BY Process_id, Created
) sq
INNER JOIN (
    SELECT Process_id, MAX(Created) AS max_created
    FROM your_table
    GROUP BY Process_id
) max_c
ON sq.Process_id = max_c.Process_id AND sq.Updated = max_c.max_created

Answer 2

除了性能之外，我就是这样做的......

SELECT a.process_id
     , a.status
     , a.created
     , b.status prev_status
     , b.created prev_created
  FROM 
     ( SELECT x.*
            , COUNT(*) rank
         FROM my_table x
         JOIN my_table y
           ON y.process_id = x.process_id
          AND y.created >= x.created
        GROUP
           BY x.process_id
            , x.created
     ) a
  LEFT
  JOIN
     ( SELECT x.*
            , COUNT(*) rank
         FROM my_table x
         JOIN my_table y
           ON y.process_id = x.process_id
          AND y.created >= x.created
        GROUP
           BY x.process_id
            , x.created
     ) b
    ON b.process_id = a.process_id
 WHERE b.rank = a.rank + 1
   AND a.rank = 1;

在更大的数据集上，我可能会选择更加精美的Pantsy解决方案

跟踪当前和以前的状态

2 个答案: