Question

VB 实时错误‘3705’ 对象打开时，不允许操作，怎样解决，求大神

代码是这样的

Dim rst As New ADODB.Recordset
Dim sql As String
Dim conn As New ADODB.Connection

Private Sub Command1_Click()
conn.ConnectionString = "provider=microsoft.jet.oledb.4.0;data source=" & App.Path & "\stu.mdb" conn.Open
If conn.State = adStateOpen Then

Print "ok"
End If
sql = "select * from stu"
rst.CursorLocation = adUseClient
rst.Open sql, conn, adOpenKeyset, adLockPessimistic
Set DataGrid1.DataSource = rst

End Sub
点一次按钮可以，点第二次就出现这种情况

Answer 1

Dim rst As New ADODB.Recordset
Dim sql As String
Dim conn As New ADODB.Connection
Private Sub Command1_Click()
If conn.State = adStateOpen Then     conn.close
conn.ConnectionString = "provider=microsoft.jet.oledb.4.0;data source=" & App.Path & "\stu.mdb"
conn.Open
sql = "select * from stu"
rst.CursorLocation = adUseClient
if rst.state=1 then rst.close
rst.Open sql, conn, adOpenKeyset, adLockPessimistic
Set DataGrid1.DataSource = rst
End Sub

Answer 2

把下面的代码注释起来
conn.Open
If conn.State = adStateOpen Then

Print "ok"
End If

追问

这样不行啊

追答

你确定把conn.Open注释了？

追问

是的，都注释了。楼下的可以，还是谢谢你。