ACC: Sample Function to Return a Random Record (108435)

SUMMARY

Microsoft Access does not have a built-in mechanism for returning a random record from a set of records. This article describes a sample user-defined function that you can use to return a random record.

This article assumes that you are familiar with Visual Basic for Applications and with creating Microsoft Access applications using the programming tools provided with Microsoft Access. For more information about Visual Basic for Applications, please refer to your version of the "Building Applications with Microsoft Access" manual.

NOTE: Visual Basic for Applications is called Access Basic in Microsoft Access versions 1.x and 2.0. For more information about Access Basic, please refer to the "Introduction to Programming" manual in Microsoft Access version 1.x or the "Building Applications" manual in Microsoft Access version 2.0.

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The following sample function will return a random record using the recordset name and the field name that you provide.

NOTE: In the following sample code, an underscore (_) at the end of a line is used as a line-continuation character. Remove the underscore from the end of the line when re-creating this code in Access Basic.

Create a module and type the following line in the Declarations section if it is not already there:
Option Explicit

Type the following procedure:

In Microsoft Access 2.0, 7.0 and 97:

      Function FindRandom (RecordSetName As String, Fieldname As String)
      Dim MyDB As Database
      Dim MyRS As Recordset
      Dim SpecificRecord As Long, i As Long, NumOfRecords As Long

      Set MyDB = CurrentDB()
      'change DB_Open_Dynaset to DBOpenDynaset for Access 97
      Set MyRS = MyDB.OpenRecordset(RecordSetName, DB_Open_Dynaset)
      On Error GoTo NoRecords
      MyRS.MoveLast
      NumOfRecords = MyRS.RecordCount
      SpecificRecord = Int(NumOfRecords * Rnd)
      If SpecificRecord = NumOfRecords Then
         SpecificRecord = SpecificRecord - 1
      End If
      MyRS.MoveFirst
      For i = 1 To SpecificRecord
          MyRS.MoveNext
      Next i
      FindRandom = MyRS(Fieldname)
      Exit Function
      NoRecords:
         If Err = 3021 Then
            MsgBox "There Are No Records In The Dynaset", 16, "Error"
         Else
            MsgBox "Error - " & Err & Chr$(13) & Chr$(10) & Error, _
               16, "Error"
         End If
      FindRandom = "No Records"
      Exit Function
      End Function

   In Microsoft Access 1.x:

      Function FindRandom (RecordSetName As String, Fieldname As String)
      Dim MyDB As Database
      Dim MyRS As DynaSet
      Dim SpecificRecord As Long, i As Long, NumOfRecords As Long

      Set MyDB = CurrentDB()
      Set MyRs = MyDB.CreateDynaset(RecordSetName)
      On Error GoTo NoRecords
      MyRS.MoveLast
      NumOfRecords = MyRS.RecordCount
      SpecificRecord = Int(NumOfRecords * Rnd)
      If SpecificRecord = NumOfRecords Then
         SpecificRecord = SpecificRecord - 1
      End If
      MyRS.MoveFirst
      For i = 1 To SpecificRecord
          MyRS.MoveNext
      Next i
      FindRandom = MyRS(Fieldname)
      Exit Function
      NoRecords:
         If Err = 3021 Then
            MsgBox "There Are No Records In The Dynaset", 16, "Error"
         Else
            MsgBox "Error - " & Err & Chr$(13) & Chr$(10) & Error, _
               16, "Error"
         End If
      FindRandom = "No Records"
      Exit Function
      End Function

To test this function, type the following line in the Debug window (or the Immediate window in versions 1.x and 2.0), and then press ENTER.
?FindRandom("<RecordSetName>", "<FieldName>")
where <RecordSetName> is the name of your recordset and <FieldName> is the name of a field in your recordset.

Note that each time you run the function, a different record will be returned.

ACC: Sample Function to Return a Random Record (108435)

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