Question

我必须将我的行转换为DB2表中的列。这就是我的表的结构。

ItemID    Item    Value
---------------------
1     Meeting     Now
1     Advise      Yes
1     NoAdvise    No
2     Meeting     Never
2     Advise      No
2     NoAdvise    Null
2     Combine    Yes

我希望将其转换为（注意我不想转置Combine）

ItemID    Meeting  Advise   NoAdvise 
---------------------------------------
1         Now      Yes       No
2         Never    No        Null

有点挣扎于查询，你能帮忙吗？

Answer 1

它不是很漂亮，但应该有效。 DB2没有内置的PIVOT函数，如SQL Server。

SELECT DISTINCT
     A.ItemID
    ,(SELECT value
      FROM table B
      WHERE B.ItemID = A.ItemID
        AND B.Item   = 'Meeting'
    ) AS Meeting
    ,(SELECT value
      FROM table B
      WHERE B.ItemID = A.ItemID
        AND B.Item   = 'Advise'
    ) AS Advise
    ,(SELECT value
      FROM table B
      WHERE B.ItemID = A.ItemID
        AND B.Item   = 'NoAdvise'
    ) AS NoAdvise
FROM table A

Answer 2

currently accepted answer by bhamby肯定是正确的，但值得检查if using several correlated subqueries is much slower than a single group by（提示：最有可能）：

SELECT 
  A.ItemID,
  MAX(CASE WHEN A.Item = 'Meeting'  THEN Value END) AS Meeting,
  MAX(CASE WHEN A.Item = 'Advise'   THEN Value END) AS Advise,
  MAX(CASE WHEN A.Item = 'NoAdvise' THEN Value END) AS NoAdvise
FROM A
GROUP BY A.ItemID

我认为这也有点简单

SQLFiddle（在PostgreSQL中，但也适用于DB2 LUW）

Answer 3

正如@bhamby所说，DB2没有PIVOT功能大多数情况下，我的查询只是在检索结果的方式上有所不同 - 您需要在它们上运行探查器/优化器以确保，但我相信相关的子查询可能会按行执行（可能效率较低），而不是作为集合。这不太可能是小数据集的问题。

WITH Item (id) as (SELECT DISTINCT itemId
                   FROM YourTable),
SELECT item.id, Meeting.meeting, Advise.advise, NoAdvise.noadvise
FROM Item
LEFT JOIN (SELECT itemId, value as meeting
           FROM YourTable
           WHERE item = 'Meeting') as Meeting
       ON Meeting.itemId = Item.id
LEFT JOIN (SELECT itemId, value as advise
           FROM YourTable
           WHERE item = 'Advise') as Advise
       ON Advise.itemId = Item.id
LEFT JOIN (SELECT itemId, value as noadvise
           FROM YourTable
           WHERE item = 'NoAdvise') as NoAdvise
       ON NoAdvise.itemId = Item.id

（...实际上，我有点担心你有'建议'和'没有建议'的列，这似乎是某种布尔条件 - 即，你应该有一个，但不是另一个）。

在DB2中进行透视

3 个答案: