Question

目前有sql返回结果集如下

WORKFLOWID  UNMATCHEDVALUE  MATCHEDADDRESS  EXCEPTIONREASON
1001        UNIQUE          ADDRESS1        (null)
1001        UNIQUE          ADDRESS2        Some Value

我正在寻找的是像这样的结果

WORKFLOWID UNMATCHEDVALUE MATCHEDADDRESS EXCEPTIONREASON MATCHEDADDRESS2    EXCEPTIONREASON2
 1001      UNIQUE         ADDRESS1       (null)          ADDRESS2           Some Value

因此“variant”列是MatchedAddress和Exception Reason，其他列对于每条记录都是相同的。请注意，对于每个workflow_id，将始终返回2行。

我还创建了一个显示模式的小提琴。

http://sqlfiddle.com/#!6/f7cde/3

Answer 1

试试这个：

;WITH CTE AS
(
    SELECT  ws.id as WorkflowStepId,
            ws.workflow_id as WorkflowId, 
            sg.unmatchValue as UnmatchedValue,
            geo_address as MatchedAddress, 
            ws.exception_Value as ExceptionReason,
            ROW_NUMBER() OVER(PARTITION BY ws.workflow_id ORDER BY ws.id) as RN
    FROM workflow_step as ws 
    INNER JOIN workflow as gw 
        ON ws.workflow_id = gw.id
    INNER JOIN super_group as sg 
        ON gw.super_group_id = sg.id
    INNER JOIN alias on 
        ws.id = alias.workflow_step_id
)
SELECT  WorkflowId,
        UnmatchedValue,
        MIN(CASE WHEN RN = 1 THEN MatchedAddress END) MatchedAddress,
        MIN(CASE WHEN RN = 1 THEN ExceptionReason END) ExceptionReason,
        MIN(CASE WHEN RN = 2 THEN MatchedAddress END) MatchedAddress2,
        MIN(CASE WHEN RN = 2 THEN ExceptionReason END) ExceptionReason2
FROM CTE
GROUP BY WorkflowId,
         UnmatchedValue
ORDER BY workflowId

Here is修改后的sqlfiddle。

结果是：

╔════════════╦════════════════╦════════════════╦═════════════════╦═════════════════╦══════════════════╗
║ WORKFLOWID ║ UNMATCHEDVALUE ║ MATCHEDADDRESS ║ EXCEPTIONREASON ║ MATCHEDADDRESS2 ║ EXCEPTIONREASON2 ║
╠════════════╬════════════════╬════════════════╬═════════════════╬═════════════════╬══════════════════╣
║       1001 ║ UNIQUE         ║ ADDRESS1       ║ (null)          ║ ADDRESS2        ║ Some Value       ║
╚════════════╩════════════════╩════════════════╩═════════════════╩═════════════════╩══════════════════╝

Answer 2

试试这个：

SELECT ws.workflow_id as WorkflowId, sg.unmatchValue as UnmatchedValue,
    MAX(CASE WHEN ws.id = 1 THEN geo_address END) as MatchedAddress1,
    MAX(CASE WHEN ws.id = 2 THEN geo_address END) as MatchedAddress2,
    MAX(CASE WHEN ws.id = 1 THEN ws.exception_Value END) as ExceptionReason1,
    MAX(CASE WHEN ws.id = 2 THEN ws.exception_Value END) as ExceptionReason2 

    FROM workflow_step as ws 
 INNER JOIN workflow as gw on ws.workflow_id = gw.id
    INNER JOIN super_group as sg on gw.super_group_id = sg.id
    inner JOIN alias on ws.id = alias.workflow_step_id
    GROUP BY  ws.workflow_id, sg.unmatchValue

SQL FIDDLE DEMO

Answer 3

由于我无法评论，我只是想指出，Lamak给出的答案是使用Common Table Expression。这些通常是解决sql中递归问题的最佳选择。

Answer 4

这假设您只有2种地址类型。如果您有更多，我建议您创建一个数据透视表。

    select a.*, MATCHEDADDRESS2,EXCEPTIONREASON2
    from
    (Select WORKFLOWID,UNIQUEVALUE,MATCHEDADDRESS,EXCEPTIONREASON
     from "Your Table"
     where MATCHEDADDRESS='ADDRESS1') a

    join

    (Select WORKFLOWID,UNIQUEVALUE,MATCHEDADDRESS as MATCHEDADDRESS2,EXCEPTIONREASON as   XCEPTIONREASON2
     from "Your Table"
     where MATCHEDADDRESS='ADDRESS2') b
          on a.WORKFLOWID=b.WORKFLOWID
               and a.UNMATCHEDVALUE = b.UNMATCHEDVALUE

从同一行的内部联接差异中获取结果

4 个答案: