继续求援!
我的数据库为ACCESS 2000。文件名为db1,密码为“1111”,路径为APP.Path
代码为:
Private Sub Command1_Click()
Dim cnData As ADODB.Connection
set cnData=new ADODB.Connection
cnData.ConnectionString = ""
cnData.Open
End Sub
如何才能在OPEN时不出错?
问题点数:40、回复次数:5Top
1 楼gxingmin(小高)回复于 2002-05-24 13:28:16 得分 10
Dim cnData As ADODB.Connection
set cnData=new ADODB.Connection
cnData.ConnectionString = "driver={Microsoft access driver (*.mdb)};dbq=" & App.Path & "\db1.MDB";pwd=1111"
cnData.ConnectionTimeout = 50
cnData.Open
Top
2 楼azure711()回复于 2002-05-24 13:37:39 得分 10
Public DataBase2 As New ADODB.Connection
Public TempRecord2 As New ADODB.Recordset
------------------------------------------------------------
DataBase2.CursorLocation = adUseClient
DataBase2.Provider = "Microsoft.Jet.OLEDB.3.51"
DataBase2.Mode = adModeUnknown
DataBase2.ConnectionString = "User ID=1111;Data Source=" & App.Path & "\db1.MDB"
DataBase2.Open
TempRecord2.LockType = adLockBatchOptimistic
TempRecord2.CursorType = adOpenKeyset
TempRecord2.Open "select * from XXXXX", DataBase2
Top
3 楼ferrytang(欢迎你)回复于 2002-05-24 13:38:09 得分 10
"Provider=Microsoft.Jet.OLEDB.4.0;Data Source="& app.path &"/db1;Persist Security Info=False;Jet OLEDB:Database Password=1111"Top
4 楼lihonggen0(李洪根,MS MVP,标准答案来了)回复于 2002-05-24 13:44:43 得分 5
Public adoCNAccess As New ADODB.Connection '定义数据库的连接存放数据和代码
Public Function OpenAccess() As String
With adoCNAccess
If .State <> adStateOpen Then
.ConnectionString = "Provider=Microsoft.Jet.OLEDB.4.0;Persist Security Info=False;Data Source=" & cProgramPath & "Trade.mdb" ';password=allway"
.ConnectionTimeout = 5
.Open
If .State = adStateOpen Then
OpenAccess = "数据库连接成功"
Else
OpenAccess = "数据库连接失败,请按帮助进行检查 !"
MsgBox "数据库连接失败,请找系统管理员进行检查 !", 16, cProgramName
End
End If
End If
End With
End Function
Top
5 楼lihonggen0(李洪根,MS MVP,标准答案来了)回复于 2002-05-24 13:55:14 得分 5
将上面的连接字符写成你的就行
.ConnectionString =Top




