Question

我有一个 Ledger Table ，其中包含以下三列以及其他未包含的内容：

**gl_account id** | **gl_amount** |  **GL_adjustmentAmount**
    9500          |     NULL      |        NULL
     ...          |   ...         |        ...

我有一个不同的调整表，上面有不同的列，除了两个：

**gl_account id** |   **GL_adjustmentamount**
     9500         |           289.84
     9500         |           9.63
     9500         |           13646.11
     9500         |           835.31
     9500         |           -210
     9500         |           -1019.02
     9500         |           -200

我需要更新 Ledger Table 以包含所有7个新gl_adjustments，但它只包含其中一个值289.84。

这是我的代码。

UPDATE LedgerTable
SET [GL_adjustmentamount] = adjust.GL_adjustmentamount 
FROM LedgerTable AS genLed
     FULL OUTER JOIN AdjustmentTable AS adjust
     ON genLed.gl_accountid = adjust.gl_accountid

Answer 1

案例1：仅针对一个帐户进行更新

如果您只想在一个帐户中使用此功能，则可以使用此功能：

declare @AccountID INT
set @AccountID = 9500

update  Ledger
set     GL_adjustmentAmount = GL_adjustmentAmount + (select     SUM(a.GL_AdjustmentAmount)
                                                     from       Adjustment a
                                                     where      a.GL_AccountID = @AccountID)
where   GL_AccountID = @AccountID

请注意，此处必须指定accountId。

案例2：所有帐户的更新

如果您希望这适用于所有帐户（应该＆amp;更可能是这种情况），那么您将需要更多＆＃34;概括＆＃34;查询类似的东西：

update Ledger
set     GL_AdjustmentAmount = ISNULL(GL_AdjustmentAmount,0) + ISNULL(collatedAdjustments.adjustment, 0)
from Ledger ledger
inner join 
(select         a.GL_AccountID,
                SUM(a.GL_AdjustmentAmount) as Adjustment
from            Adjustment adjustments
group by        adjustments.GL_accountID) as collated
on              ledger.GL_accountID = collated.GL_accountID

这里有一些示例数据可供您测试：

CREATE TABLE Ledger(GL_AccountID int, GL_Amount int, GL_AdjustmentAmount int);
CREATE TABLE Adjustment(GL_AccountID int, GL_AdjustmentAmount int);

INSERT INTO Ledger(GL_AccountID,GL_Amount, GL_AdjustmentAmount) VALUES(9500, null, null);
INSERT INTO Ledger(GL_AccountID,GL_Amount, GL_AdjustmentAmount) VALUES(9600, null, null);
INSERT INTO Ledger(GL_AccountID,GL_Amount, GL_AdjustmentAmount) VALUES(9700, null, null);
INSERT INTO Ledger(GL_AccountID,GL_Amount, GL_AdjustmentAmount) VALUES(9800, null, null);

INSERT INTO Adjustment(GL_AccountID, GL_AdjustmentAmount) VALUES(9500, 289.84);
INSERT INTO Adjustment(GL_AccountID, GL_AdjustmentAmount) VALUES(9500, 9.63);
INSERT INTO Adjustment(GL_AccountID, GL_AdjustmentAmount) VALUES(9500, 13646.11);
INSERT INTO Adjustment(GL_AccountID, GL_AdjustmentAmount) VALUES(9500, 835.31);
INSERT INTO Adjustment(GL_AccountID, GL_AdjustmentAmount) VALUES(9500, -210);
INSERT INTO Adjustment(GL_AccountID, GL_AdjustmentAmount) VALUES(9500, -1019.02);
INSERT INTO Adjustment(GL_AccountID, GL_AdjustmentAmount) VALUES(9500, -200);

INSERT INTO Adjustment(GL_AccountID, GL_AdjustmentAmount) VALUES(9600, 29.84);
INSERT INTO Adjustment(GL_AccountID, GL_AdjustmentAmount) VALUES(9600, 29.63);
INSERT INTO Adjustment(GL_AccountID, GL_AdjustmentAmount) VALUES(9600, 16646.11);
INSERT INTO Adjustment(GL_AccountID, GL_AdjustmentAmount) VALUES(9600, 335.31);
INSERT INTO Adjustment(GL_AccountID, GL_AdjustmentAmount) VALUES(9600, -1210);
INSERT INTO Adjustment(GL_AccountID, GL_AdjustmentAmount) VALUES(9600, -2019.02);
INSERT INTO Adjustment(GL_AccountID, GL_AdjustmentAmount) VALUES(9600, -1200);

INSERT INTO Adjustment(GL_AccountID, GL_AdjustmentAmount) VALUES(9700, 2239.02);
INSERT INTO Adjustment(GL_AccountID, GL_AdjustmentAmount) VALUES(9700, 1400);

INSERT INTO Adjustment(GL_AccountID, GL_AdjustmentAmount) VALUES(9800, 4121.02);
INSERT INTO Adjustment(GL_AccountID, GL_AdjustmentAmount) VALUES(9800, 1234);

此查询也会将之前的任何调整（在ledger表中）纳入等式中。

希望这有帮助!!!

Answer 2

select * into LedgerTable
from Adjustment Table

尝试使用另一个表更新一个表

2 个答案: