Question

我需要从select中插入行，如果key是重复的，则删除条目并插入

 Table1 

  ColumnA   ColumnB   ColumnC  ColumnD
   A          1        A1        7/21/2017
   B          2        B1        7/22/2017
   C           3       C1       7/23/2017

独特的组合ColumnA和ColumnC

 Table2

  ColumnE  ColumnF ColumnG
    A        1      A1
    A         2     A2
    B         3     B1
    B         2     B2
    C         3     C1
    C         1     C2

我应该将表2中的行插入表1

Insert into table1 (columnA, columnB, ColumnC) select columnE, ColumnF, ColumnG from table2

上述查询会出现冲突，说明插入了重复的密钥但需要一种删除创建冲突并插入该行的行的方法。最终输出应该是

 Table 1
 ColumnA  ColumnB ColumnC  ColumnD  
    A        1      A1     08/08/2017  - deleted and added as conflict arised
    A         2     A2     08/08/2017
    B         3     B1     08/08/2017  -deleted and added as conflict arised
    B         2     B2     08/08/2017
    C         3     C1     08/08/2017    
    C         1     C2     08/08/2017

Answer 1

您不必删除然后插入。只需更新$scope.myGrid = { rowData: $scope.data, columnDefs: [...], onModelUpdated: function(event) { MyFactory.onModelUpdated(this); } };列（或所有非键列），然后从ColumnD插入新列。我会先命令语句，然后先Table2然后再UPDATE，这样INSERT就可以处理较小的原始行集（在插入之前）。除了UPDATE之外，它只是普通的MERGE：

UPDATE-INSERT

现在，在表变量DECLARE @Conflicted TABLE( ColumnA char(1), ColumnB int, ColumnC char(2), ColumnD date ) BEGIN TRAN UPDATE dst SET dst.ColumnD = GETDATE() OUTPUT deleted.ColumnA, deleted.ColumnB, deleted.ColumnC, deleted.ColumnD INTO @Conflicted FROM Table1 AS dst JOIN Table2 AS src ON dst.ColumnA = src.ColumnE AND dst.ColumnB = src.ColumnF AND dst.ColumnC = src.ColumnG INSERT INTO Table1( ColumnA, ColumnB, ColumnC, ColumnD ) SELECT t2.ColumnE, t2.ColumnF, t2.ColumnG, GETDATE() FROM Table2 AS t2 WHERE NOT EXISTS(SELECT * FROM Table1 WHERE ColumnA = t2.ColumnE AND ColumnB = t2.ColumnF AND ColumnC = t2.ColumnG) COMMIT TRAN中，@Conflicted中的所有冲突行都已被Table1中的新行替换。

这是一个Table2的版本：

MERGE

SQL中的重复键错误以及删除和更新

1 个答案: