Question

我正在尝试将数据库中的字段更新为其连接值的总和：

UPDATE P
SET extrasPrice = SUM(E.price)
FROM dbo.BookingPitchExtras AS E
INNER JOIN dbo.BookingPitches AS P ON E.pitchID = P.ID
    AND P.bookingID = 1
WHERE E.[required] = 1

当我运行时，我收到以下错误：

"An aggregate may not appear in the set list of an UPDATE statement."

有什么想法吗？

Answer 1

这个怎么样：

UPDATE p
SET p.extrasPrice = t.sumPrice
FROM BookingPitches AS p
INNER JOIN
    (
        SELECT PitchID, SUM(Price) sumPrice
        FROM BookingPitchExtras
        WHERE [required] = 1
        GROUP BY PitchID 
    ) t
    ON t.PitchID = p.ID
WHERE p.bookingID = 1

Answer 2

上述解决方案的替代方案是使用表的别名：

UPDATE T1 SET T1.extrasPrice = (SELECT SUM(T2.Price) FROM BookingPitchExtras T2 WHERE T2.pitchID = T1.ID)
FROM BookingPitches T1;

Answer 3

这是一个有效的错误。见this。以下（以及其他建议如下）是实现这一目标的方法： -

UPDATE P 
SET extrasPrice = t.TotalPrice
FROM BookingPitches AS P INNER JOIN
 (
  SELECT
    PitchID,
    SUM(Price) TotalPrice
  FROM
     BookingPitchExtras
  GROUP BY PitchID
  ) t
ON t.PitchID = p.ID

Answer 4

我遇到了同样的问题，发现我可以用Common Table Expression解决它（在SQL 2005或更高版本中可用）：

;with cte as (
    SELECT PitchID, SUM(Price) somePrice
    FROM BookingPitchExtras
    WHERE [required] = 1 
    GROUP BY PitchID)
UPDATE p SET p.extrasPrice=cte.SomePrice
FROM BookingPitches p INNER JOIN cte ON p.ID=cte.PitchID
WHERE p.BookingID=1

Answer 5

使用postgres，我不得不调整解决方案，以便为我工作：

UPDATE BookingPitches AS p
SET extrasPrice = t.sumPrice
FROM 
    (
        SELECT PitchID, SUM(Price) sumPrice
        FROM BookingPitchExtras
        WHERE [required] = 1
        GROUP BY PitchID 
    ) t
WHERE t.PitchID = p.ID AND p.bookingID = 1

Answer 6

使用类似于下面的子查询。

UPDATE P
SET extrasPrice = sub.TotalPrice from
BookingPitches p
inner join 
(Select PitchID, Sum(Price) TotalPrice
    from  dbo.BookingPitchExtras
    Where [Required] = 1
    Group by Pitchid
) as Sub
on p.Id = e.PitchId 
where p.BookingId = 1

Answer 7

你需要这样的东西：

UPDATE P
SET ExtrasPrice = E.TotalPrice
FROM dbo.BookingPitches AS P
INNER JOIN (SELECT BPE.PitchID, Sum(BPE.Price) AS TotalPrice
    FROM BookingPitchExtras AS BPE
    WHERE BPE.[Required] = 1
    GROUP BY BPE.PitchID) AS E ON P.ID = E.PitchID
WHERE P.BookingID = 1

SQL更新其连接值的SUM

7 个答案: