ACC: New SQL Records Appear Deleted Until Recordset Reopened

• Microsoft Access versions 7.0, 97

SYMPTOMS

Advanced: Requires expert coding, interoperability, and multiuser skills.

When you add a record to an SQL table by using Visual Basic for Applications, if the table's unique index field has a default value, and you do not assign a value to that field, the new record appears deleted until you reopen the SQL table. If you try to obtain a value from the new record, you receive the following error message:

Run-time error '3167'
Record is deleted.

RESOLUTION

When you open the SQL table by using Visual Basic code, include the dbSeeChanges option, as in the following example:

   Set rs = db.OpenRecordset("TestTable", dbOpenDynaset, dbSeeChanges) 

STATUS

This behavior is by design.

MORE INFORMATION

Steps to Reproduce Behavior

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

   
         Option Explicit 




2. Type the following procedure:
   

   
         Function TestSQLData()
   
            Dim db As Database, rs As Recordset
            Dim idx, td
            Dim cmd As String
   
            ' Delete TestTable if it exists on the SQL server.
            Set db = OpenDatabase("", False, False,ODBC;dsn=<datasource>; _
               database=<database>;uid=<user id>;pwd=<password>")
            cmd = "if exists (select * from sysobjects where _
               id = object_id('dbo.TestTable'))"
            cmd = cmd & " drop table TestTable"
            db.Execute cmd, dbSQLPassThrough
   
            ' Create TestTable with one field on SQL server.
            Set td = db.CreateTableDef("TestTable")
            td.Fields.Append td.CreateField("Int", dbInteger)
            td.Fields.Append td.CreateField("String", dbText, 50)
            db.TableDefs.Append td
   
            Set idx = td.CreateIndex("MyIdx")
            idx.Unique = True
            idx.Fields.Append idx.CreateField("Int")
            td.Indexes.Append idx
   
            cmd = "create Default TestDef3 as 100"
            db.Execute cmd, dbSQLPassThrough
   
            cmd = "sp_bindefault TestDef3, 'TestTable.Int'"
            db.Execute cmd, dbSQLPassThrough
   
            ' Open table, add a record, and then obtain values.
            Set rs = db.OpenRecordset("TestTable")
            rs.AddNew
            rs!String = "Trial"
            rs.Update
   
            Debug.Print "RecordCount = " & rs.RecordCount
            rs.MoveFirst
            Debug.Print "String is " & rs("String")
            Debug.Print "Int is " & rs("Int")
            rs.Close
   
         End Function 




3. To test this function, type the following line in the Debug window, and then press ENTER:
   
   ? TestSQLData()
   
   Note that run-time error '3167' occurs.

REFERENCES

For more information about the OpenRecordset method, search the Help Index for "OpenRecordset," and then "OpenRecordset method," or ask the Microsoft Access 97 Office Assistant.

Keywords          : kberrmsg kbusage OdbcProb 
Version           : 7.0 97
Platform          : WINDOWS 
Issue type        : kbprb 

Last Reviewed: April 23, 1999