Question

查看以下以下代码行。

Dim rst As DAO.Recordset
Dim strSql As String

strSql = "SELECT * FROM MachineSettingsT;"
Set rst = DBEngine(0)(0).OpenRecordset(strSql)

rst.FindFirst "Microwave = " & "'" & Me.Microwave & "'" & " AND WashingMachine =" & "'" & Me.WashingMachine & "'" & " AND Element1 =" & "'" & Me.Element1 & "'" _
               & "AND Element3 =" & "'" & Me.Element3 & "'" & "AND Dryer =" & "'" & Me.Dryer & "'" & "AND SettingID <>" & "'" & Me.SettingID & "'"

If Not rst.NoMatch Then  
    Cancel = True
    If MsgBox("Setting already exists; go to existing record?", vbYesNo) = vbYes Then
        Me.Undo
        DoCmd.SearchForRecord , , acFirst, "[SettingID] = " & rst("SettingID")
    End If
End If
rst.Close

问题：如果rst.FindFirst表达式中的任何值为Null，则即使在正在评估的字段中存在匹配的Null值的记录时，rst.NoMatch也始终返回true。这种行为是预期的还是会有另一个潜在的问题。我检查了msdn页面，但没有提供有关此类行为的信息。

Answer 1

考虑一种不同的方法。请注意，这是针对一组文本数据类型字段的。

Dim rst As DAO.Recordset
Dim strSql As String
Dim db As Database

Set db=CurrentDB

strSql = "SELECT * FROM MachineSettingsT WHERE 1=1 "
''Set rst = db.OpenRecordset(strSql)

If not IsNull(Me.Microwave) Then
   strWhere =  " AND Microwave = '" & Me.Microwave & "'" 
End if
If not IsNull(Me.WashingMachine) Then
   strWhere = strWhere & " AND WashingMachine ='" & Me.WashingMachine & "'"
End if
If not IsNull(Me.Element1) Then
   strWhere = strWhere & " AND Element1 ='" & Me.Element1 & "'" 
End if
If not IsNull(Me.Element3) Then
   strWhere = strWhere & " AND Element3 ='" & Me.Element3 & "'"  
End if
If not IsNull(Me.Dryer) Then
   strWhere = strWhere & " AND Dryer ='" & Me.Dryer & "'"
End if

Set rst = db.OpenRecordset(strSql & strWhere)

Answer 2

当控件值为空时，您的.FindFirst 条件必须检查相应的字段Is Null是否等于控件的值。从一个更简单的例子开始，检查两个控制/字段对。

Dim strCriteria As String
If IsNull(Me.Microwave) Then
    strCriteria = " AND Microwave Is Null"
Else
    strCriteria = " AND Microwave = '" & Me.Microwave & "'"
End If
If IsNull(Me.WashingMachine) Then
    strCriteria = strCriteria & " AND WashingMachine Is Null"
Else
    strCriteria = strCriteria & " AND WashingMachine = '" & Me.WashingMachine & "'"
End If
If Len(strCriteria) > 0 Then
    ' discard leading " AND "
    strCriteria = Mid(strCriteria, 6)
    Debug.Print strCriteria 
    rst.FindFirst strCriteria
End If

.FindFirst VBA MS Access函数的意外行为：.NoMatch始终返回true

2 个答案: