表连接以获得结果

Question

这是需要更新的表。

TABLE A
-------------------
ID   UserID   Value
-------------------
1    1        1A
2    1        1B
3    1        1C
4    2        2A
5    3        3A
6    4        4A

我有一个临时表，其中包含已更新用户的新值。

TEMP TABLE
-------------
UserID  Value
-------------
1   1A         --existing
1   1D         --new
2   2B         --new

我想知道如何更新 TABLE A 以反映 TEMP TABLE 中的新值。预期的结果是：

TABLE A
-------------------
ID   UserID   Value
-------------------
1    1        1A
7    1        1D
8    2        2B
5    3        3A
6    4        4A

我有两个想法：

• 先删除再插入。使用左连接确定哪些值不再存在，然后将其删除。然后使用右连接确定哪些值是新值，然后插入它们。
• 也许我可以使用MERGE，但是我很确定如何实现它。

测试环境

IF OBJECT_ID('tempdb..#tableA') IS NOT NULL
BEGIN
    DROP TABLE #tableA
END

IF OBJECT_ID('tempdb..#tempTable') IS NOT NULL
BEGIN
    DROP TABLE #tempTable
END

CREATE TABLE #tableA
(
    ID int identity(1,1),
    UserID int,
    Value nvarchar(50)
)

CREATE TABLE #tempTable
(
    UserID int,
    Value nvarchar(50)
)

INSERT INTO #tableA([UserID], [Value])
VALUES (1, '1A'),
(1, '1B'),
(1, '1C'),
(2, '2A'),
(3, '3A'),
(4, '4A')

INSERT INTO #tempTable([UserID], [Value])
VALUES (1, '1A'),
(1, '1D'),
(2, '2B')

SELECT * FROM #tableA
SELECT * FROM #tempTable

编辑：以下解决方案从表 A 中删除 ID (1,2,3,4)，但我只希望它删除 (2,3,4)。这是因为 ID 1 已经存在于 TABLE A 中，不需要删除再重新插入。

Delete 
From    TableA  A
Where Exists
(
    Select  *
    From    TempTable   T
    Where   T.UserId = A.UserId
)

我提供了一个解决方案，但是我发现它很混乱。有没有办法让它变得更好？

-- This will get the IDS (1,2,3,4) from TABLE A
SELECT * INTO #temp1 FROM #tableA
WHERE EXISTS
(
    Select *
    From #tempTable T
    Where T.UserId = #tableA.UserId
)

--This will get the ID (1) from TABLE A. I do not want this deleted.
SELECT * INTO #temp2 FROM #tableA
WHERE EXISTS
(
    Select *
    From #tempTable T
    Where T.UserId = #tableA.UserId AND T.[Value]=#tableA.Value
)

--LEFT JOIN to only delete the IDS (2,3,4)
DELETE FROM #tableA
WHERE EXISTS
(
    SELECT *
    FROM #temp1 a LEFT JOIN #temp2 b
    ON a.UserID=b.UserID AND a.Value=b.Value
    WHERE b.UserID IS NULL AND b.Value IS NULL
)

Answer 1

--This will remove all records which have at least one row with a matching UserID in tempTable and which don't have a row that matches on both UserID and Value.
DELETE
  TableA  A
WHERE 
 A.UserID IN (SELECT DISTINCT USerID FROM TempTable)
AND NOT EXISTS
(
    Select  1
    From    TempTable   T
    Where   T.UserId = A.UserId
    AND     T.Value= A.Value
)
--This will add any rows from temptable that don't have a match already in TableA
INSERT INTO TableA(UserId, Value)
SELECT DISTINCT UserID,Value
FROM TempTable T
WHERE NOT EXISTS (SELECT 1 FROM TableA A 
WHERE T.UserID=A.UserID
AND T.value=A.Value)

这会得到你想要的结果。如果它是一个巨大的结果集 - 无论是现实生活中更宽的表还是数百万行，那么可能会影响性能，需要后退一步。否则，会成功的

Answer 2

由于您要删除临时表中未出现的记录，我将使用删除/插入方法：

Delete 
From    TableA  A
Where Exists
(
    Select  *
    From    TempTable   T
    Where   T.UserId = A.UserId
)

Insert
Into    TableA
        (UserId, Value)
Select  T.UserId, T.Value
From    TempTable   T
Where Not Exists
(
    Select  *
    From    TableA  A
    Where   A.UserId = T.UserId
    And     A.Value = T.Value
)

您也可以使用这些的Outer Join 方法：

Delete  A
From    TableA      A
Join    TempTable   T   On  T.UserId = A.UserId

Insert
Into        TableA
            (UserId, Value)
Select      T.UserId, T.Value
From        TempTable   T
Left Join   TableA      A   On  A.UserId = T.UserId
                            And A.Value = T.Value
Where       A.UserId Is Null

就我个人而言，Right Joins 只是伤了我的脑筋，所以请随意更改顺序。