Article ID: 210468 - Last Review: October 11, 2006 - Revision: 2.2

ACC2000: Sample Function to Return a Random Record from a Field

View products that this article applies to.

This article was previously published under Q210468

This article applies only to a Microsoft Access database (.mdb).

Moderate: Requires basic macro, coding, and interoperability skills.

Expand all | Collapse all

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.

Microsoft provides programming examples for illustration only, without warranty either expressed or implied. This includes, but is not limited to, the implied warranties of merchantability or fitness for a particular purpose. This article assumes that you are familiar with the programming language that is being demonstrated and with the tools that are used to create and to debug procedures. Microsoft support engineers can help explain the functionality of a particular procedure, but they will not modify these examples to provide added functionality or construct procedures to meet your specific requirements.

Back to the top

MORE INFORMATION

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

NOTE: The sample code in this article uses Microsoft Data Access Objects. For this code to run properly, you must reference the Microsoft DAO 3.6 Object Library. To do so, click References on the Tools menu in the Visual Basic Editor, and make sure that the Microsoft DAO 3.6 Object Library check box is selected.

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

Option Explicit

Type the following procedure:

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()
   Set MyRS = MyDB.OpenRecordset(RecordSetName, dbOpenDynaset)
   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 Immediate window, and then press ENTER:

?FindRandom("<RecordSetName>", "<FieldName>")

where <RecordSetName> is the name of a table or query or a SQL statement and <FieldName> is the name of a field in your recordset.

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

Back to the top

REFERENCES

For additional information about using a query to return random records, please see the following articles in the Microsoft Knowledge Base:

208855 (http://support.microsoft.com/kb/208855/EN-US/ ) ACC2000: Find N Records in Random Order

For more information about <B<Int() function or the <Bnd() function, click Microsoft Access 2000 Help on the Help menu, type Int or Rnd in the Office Assistant or the Answer Wizard, and then click Search to view the topics returned.

Back to the top