Question

我有2张桌子

Delivery
--------
deliveryid int (PK)

description long varchar

DeliveryHistory
---------------
historyid int

delievryid int

statusid int

recordtime timestamp

我想要做的是左外连接以从表交付中带回所有记录，每个交付只有DeliveryHistory中的最新条目。但是，如果DeliveryHistory中没有用于交付的条目，我想要一个空值

我做到了：

select d.deliveryid,d.description, h.statusid from delivery d
left outer join  Deliveryhistory h on d.deliveryid = h.deliveryid
where  h.recordtime =
       ( SELECT MAX(recordtime)
           FROM Deliveryhistory
          WHERE deliveryid = d.deliveryid)

但它只返回在DeliveryHistory中有条目的行。

Answer 1

您的where子句导致排除所有空值。试试

where  h.RecordTime is null OR
       h.recordtime =
       ( SELECT MAX(recordtime)
           FROM Deliveryhistory
          WHERE deliveryid = d.deliveryid)

Answer 2

select d.deliveryid,d.description, h.statusid from delivery d
left outer join  Deliveryhistory h on d.deliveryid = h.deliveryid
where  (h.recordtime =
   ( SELECT MAX(recordtime)
       FROM Deliveryhistory
      WHERE deliveryid = d.deliveryid)
  or h.deliveryid = null)

Answer 3

现有的答案都是如此，但是如果你不想使用WHERE子句就可以这样做，你可以使用以下构造。

SELECT  d.deliveryid
        ,d.description
        , dh.statusid
FROM    Delivery d 
        LEFT OUTER JOIN (
          SELECT deliveryid, MAX(recordtime) AS recordtime
          FROM   DeliveryHistory
          GROUP BY
                 deliveryid
        ) dhm ON dhm.deliveryid = d.deliveryid                 
        LEFT OUTER JOIN DeliveryHistory dh ON dh.deliveryid = dhm.deliveryid 
                                              AND dh.recordtime = dhm.recordtime

Answer 4

CTE产生maxrow（IFF实现支持CTE ;-)加上简单的左连接与CTE。

WITH last AS (
        SELECT * FROM Deliveryhistory dh
        WHERE NOT EXISTS (
                SELECT * 
                FROM Deliveryhistory nx
                WHERE nx.deliveryid = dh.deliveryid
                AND nx.recordtime > dh.recordtime -- no one is bigger: dh must be the max
                )
        )
SELECT d.deliveryid, d.description, l.statusid 
FROM delivery d
LEFT JOIN last l ON d.deliveryid = l.deliveryid
        ;

在第二张桌子上加入最近记录的记录

4 个答案: