Vba 在Access中将一行从一个表复制到另一个表

Vba 在Access中将一行从一个表复制到另一个表,vba,ms-access,Vba,Ms Access,这是我第一次使用Access,所以我现在有点困惑。这是我的代码,我不知道哪部分错了。没有错误,但单击按钮后没有发生任何事情。谢谢代码如下: Private Sub Command12_Click() Dim db As Database Dim rs As DAO.Recordset Dim rs2 As DAO.Recordset Set db = CurrentDb() Set rs = db.OpenRecordset("Amity") Set rs2 = db.OpenRecordse

这是我第一次使用Access,所以我现在有点困惑。这是我的代码,我不知道哪部分错了。没有错误,但单击按钮后没有发生任何事情。谢谢代码如下:

Private Sub Command12_Click()

Dim db As Database
Dim rs As DAO.Recordset
Dim rs2 As DAO.Recordset
Set db = CurrentDb()
Set rs = db.OpenRecordset("Amity")
Set rs2 = db.OpenRecordset("Opportunity")
With rs2
.AddNew
.Fields("Donor_Code") = rs!Donor_Code
.Update
.Close
End With
rs.Close

End Sub

您可以使用Do循环尝试此操作:

Private Sub Command12_Click()

  Dim db As Database
  Dim rs As DAO.Recordset
  Dim rs2 As DAO.Recordset

  Set db = CurrentDb()    
  Set rs = db.OpenRecordset("Amity")
  Set rs2 = db.OpenRecordset("Opportunity")

  Do While (Not rs.EOF)
    rs2.AddNew
    rs2.Fields("Donor_Code").Value = rs!Donor_Code.Value
    rs2.Update
    rs.MoveNext
  Loop
'
  rs2.Close
  Set rs2 = Nothing
  rs.Close
  Set rs = Nothing
  Set db = Nothing
'
End Sub

我刚刚尝试了这些命令,但仍然没有任何更改和错误:Private Sub Command12_Click()Dim db As Database Dim rs As DAO.Recordset Dim rs2 As DAO.Recordset db=CurrentDb()Set rs=db.OpenRecordset(“友好”)Set rs2=db.OpenRecordset(“机会”),而不是rs.EOF rs2.AddNew rs2![捐赠者代码]=rs![捐赠者代码]rs2.更新rs.MoveNext Wend rs2.关闭rs.关闭End SubAre您确定此代码已运行吗?尝试设置断点或检查按钮Command12是否有event.onClick属性表。如果有,它会导致你的潜艇