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I am trying to .FindLast to search for a specific record, and it was working with one criteria, but when I tried to use .FindLast with multiple criteria it stopped working. However, I use almost the same statement with .FindFirst and it works which is why I am confused.

The error I get is "Data type mismatch in criteria expression". And the error is for this line: rst.FindLast ("DONOR_CONTACT_ID= 'strDonor1' AND ORDER_NUMBER= 'strOrderNum1'"). I stepped through my code and the line .FindFirst ("DONOR_CONTACT_ID= 'strDonor1' and ORDER_NUMBER= 'strOrderNum1'") works correctly however. 

Option Compare Database
Option Explicit

Public dbs As DAO.Database
Public rst As DAO.Recordset
Public rstOutput As DAO.Recordset
'Defines DAO objects
Public strDonor1 As Variant
Public strDonor2 As Variant
Public strRecip1 As Variant
Public strRecip2 As Variant
Public strOrderNum1 As Variant
Public strOrderNum2 As Variant
Public strLastDonor As Variant

Function UsingTemps()

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
strDonor1 = rst!DONOR_CONTACT_ID
'sets strTemp1 to the first record of the DONOR_CONTACT_ID
strRecip1 = rst!RECIPIENT_CONTACT_ID
strOrderNum1 = rst!ORDER_NUMBER
rst.MoveNext
'moves to the next record

Do While Not rst.EOF
'Loop while it's not the end of the file
    strDonor2 = rst!DONOR_CONTACT_ID
    'strTemp2 = DONOR_CONTACT_ID from T_RECIPIENT_SORT
    strRecip2 = rst!RECIPIENT_CONTACT_ID
    strOrderNum2 = rst!ORDER_NUMBER
    'Sets strRecip = RECIPIENT_CONTACT_ID FROM T_RECIPIENT_SORT
    With rstOutput
    'Uses T_OUTPUT table
    If (strDonor1 = strDonor2) And (strOrderNum1 = strOrderNum2) Then
    'Runs if temps have same DONOR_CONTACT ID

            If .RecordCount > 0 Then
            'If table has records then you can check

                rst.FindLast ("DONOR_CONTACT_ID= 'strDonor1' AND ORDER_NUMBER= 'strOrderNum1'")
                strLastDonor = rst!RECIPIENT_CONTACT_ID
                If strLastDonor = strRecip2 Then
                    Call LastDonor
                Else
                    Call FirstDonor
                End If
            Else
            'No records in T_Output so needs to add first record
                .AddNew
                !DONOR_CONTACT_ID = strDonor1
                !RECIPIENT_1 = strRecip1
                !ORDER_NUMBER = strOrderNum1
                .Update
            End If
    Else
        .FindFirst ("DONOR_CONTACT_ID= 'strDonor1' and ORDER_NUMBER= 'strOrderNum1'")
        If .NoMatch Then
            .AddNew
            !DONOR_CONTACT_ID = strDonor1
            !RECIPIENT_1 = strRecip1
            !ORDER_NUMBER = strOrderNum1
            .Update
        End If

    End If
    End With
    'Slides variables down
    rst.FindFirst "[RECIPIENT_CONTACT_ID] = " & strRecip2
    strDonor1 = strDonor2
    strRecip1 = strRecip2
    strOrderNum1 = strOrderNum2
    rst.MoveNext

Loop

Call LastRecord

Set dbs = Nothing
Set rst = Nothing
Set rstOutput = Nothing

End Function

EDIT:

I just added the following code:

Dim strFind As Variant
strFind = "DONOR_CONTACT_ID= '" & strDonor1 & "' AND ORDER_NUMBER= '" & strOrderNum1 & "'"
Debug.Print strFind
rst.FindLast strFind

It displayed this with the Debug.Print: 

DONOR_CONTACT_ID= '10136851341' AND ORDER_NUMBER= '112103071441001'

These are the correct values for DONOR_CONTACT_ID and ORDER_NUMBER but I am getting the error "Data type mismatch in criteria expression" with the line rst.FindLast strFind. Could it possibly be because I defined my variables as variants? In the table I have DONOR_CONTACT_ID defined as Decimal with 11 precision, RECIPIENT_CONTACT_ID defined as Decimal with 11 precision, and ORDER_NUMBER as Decimal with 15 precision. I then define the variables in my code as variants. Do you think there could be a problem with this?

4

2 回答 2

4

我认为如果您更改此设置，您的故障排除工作会更容易......

rst.FindLast ("DONOR_CONTACT_ID= 'strDonor1' AND ORDER_NUMBER= 'strOrderNum1'")

像这样的事情......

Dim strFind As String
strFind = "DONOR_CONTACT_ID= 'strDonor1' AND ORDER_NUMBER= 'strOrderNum1'"
Debug.Print strFind
rst.FindLast strFind

当代码抛出错误，或者根本找不到您期望的内容时，转到立即窗口 ( Ctrl+ g) 并检查来自Debug.Print strFind. 您可能会立即发现问题。如果没有，复制Debug.Print输出，在查询设计器中打开一个新查询，切换到 SQL 视图并在WHERE子句中使用复制的文本。在这种情况下，我认为查询 SQL 可能是：

SELECT *
FROM T_RECIPIENT_SORT
WHERE yadda_yadda;

将yadda_yadda替换为您从立即窗口复制的文本。

这更像是一般的故障排除建议。对于这个特定问题，我认为您正在构建Find文本以包含变量的名称而不是那些变量的值。看看当你使用Debug.Print这 2 个字符串表达式时你会得到什么。

"DONOR_CONTACT_ID= 'strDonor1' AND ORDER_NUMBER= 'strOrderNum1'"
"DONOR_CONTACT_ID= '" & strDonor1 & "' AND ORDER_NUMBER= '" & strOrderNum1 & "'"

您的代码使用了第一个，但我认为您实际上需要第二个。

在您报告的问题的更新中DONOR_CONTACT_ID，ORDER_NUMBER它们都是数字数据类型。在这种情况下，不要在Find字符串中引用这些搜索值。

"DONOR_CONTACT_ID= " & strDonor1 & " AND ORDER_NUMBER= " & strOrderNum1
于 2013-08-02T16:08:36.257 回答
0

我们可以在 DONOR_CONTACT_ID 匹配但 ORDER_NUMBER 为空的地方丢失一些数据吗？我认为 Access 会抛出你从那种情况中得到的那种错误。

除非第一次出现是罪魁祸首，否则不会在 FindFirst 上发生。

于 2013-08-02T14:57:44.700 回答