vba - 如何将 rst.FindFirst 与 rst.NoMatch 一起使用？

Question

除了这一行，我的代码有效

.FindFirst "[DONOR_CONTACT_ID] = strTemp2"

我希望我的代码检查是否有记录，其中存在特定的 DONOR_CONTACT_ID，因为有多个记录具有相同的 DONOR_CONTACT_ID。如果该记录不存在，那么我想将该 DONOR_CONTACT_ID 和 RECIPIENT_CONTACT_ID 添加到 RECIPIENT_1。如果该记录确实存在，我想为该特定 DONOR_CONTACT_ID 添加 RECIPIENT_CONTACT_ID 到 RECIPIENT_2。为此，我使用了 .FindFirst 来查看是否有记录，然后使用了 .NoMatch。如果不匹配，我想添加一个新记录，但如果有，我想检查它是否必须进入 RECIPIENT_2。

我得到的错误是“无法将‘strTemp2’识别为有效的字段名称或表达式”。我想看看记录是否等于strTemp2，但我认为我的语法是错误的。谢谢你的帮助!!

这是我的代码:

Option Compare Database
Option Explicit

Function UsingTemps()

Dim dbs As DAO.Database
Dim rst As DAO.Recordset
Dim rstOutput As DAO.Recordset
'Defines DAO objects
Dim strTemp1 As String
Dim strTemp2 As String
Dim strVal As String
Dim strRecip As String

DoCmd.SetWarnings False
DoCmd.OpenQuery ("Q_RECIPIENT_SORT")
DoCmd.OpenQuery ("Q_DELETE_T_OUTPUT")
DoCmd.SetWarnings True
Set dbs = CurrentDb

Set rst = dbs.OpenRecordset("T_RECIPIENT_SORT", dbOpenDynaset)
'rst refers to the table T_RECIPIENT_SORT
Set rstOutput = dbs.OpenRecordset("T_OUTPUT", dbOpenDynaset)
'rstOutput refers to the table T_OUTPUT

rst.MoveFirst
'first record
strTemp1 = rst!DONOR_CONTACT_ID
'sets strTemp1 to the first record of the DONOR_CONTACT_ID
rst.MoveNext
'moves to the next record


    Do While Not rst.EOF
    'Loop while it's not the end of the file
        strTemp2 = rst!DONOR_CONTACT_ID
        'strTemp2 = DONOR_CONTACT_ID from T_RECIPIENT_SORT

    If strTemp1 = strTemp2 Then
    'Runs if temps have same DONOR_CONTACT ID
        strRecip = rst!RECIPIENT_CONTACT_ID
    'Sets strRecip = RECIPIENT_CONTACT_ID FROM T_RECIPIENT_SORT

        With rstOutput
        'Uses T_OUTPUT table
            If .RecordCount > 0 Then
            'If table has records then you can check
                .FindFirst "[DONOR_CONTACT_ID] = strTemp2"
                If .NoMatch Then
                    .AddNew
                    !DONOR_CONTACT_ID = strTemp1
                    !RECIPIENT_1 = strRecip
                    .Update
                Else

                    If !DONOR_CONTACT_ID = strTemp2 Then
                        If IsNull(!RECIPIENT_2) And Not (IsNull(!RECIPIENT_1)) Then
                            .Edit
                            !RECIPIENT_2 = strRecip
                            .Update
                        End If
                        .AddNew
                        !DONOR_CONTACT_ID = strTemp2
                        !RECIPIENT_1 = strRecip
                        .Update
                    End If
                End If

            Else
                .AddNew
                !DONOR_CONTACT_ID = strTemp2
                !RECIPIENT_1 = strRecip
                .Update
            End If

        End With

    End If

    strTemp1 = strTemp2
    rst.MoveNext

Loop

Set dbs = Nothing

End Function

Answer 1

将变量的值而不是变量名构建到您提供的字符串中 FindFirst找到。

假设 DONOR_CONTACT_ID是文本数据类型，还包括变量值周围的引号...

.FindFirst "[DONOR_CONTACT_ID] = '" & strTemp2 & "'"

但如果它是数字，你就不需要那些引号......

.FindFirst "[DONOR_CONTACT_ID] = " & strTemp2