The Microsoft ODBC Driver for Access and the Microsoft OLE DB Provider for
Jet do not provide support for bitwise operations in SQL statements.
Attempts to use AND, OR, and XOR with numeric fields in a SQL statement
return the result of a logical operation (true or false).
The Microsoft Jet database engine does not support bitwise operations in
SQL.
Bitwise operations must be replaced with the equivalent mathematical
expressions or performed on the data outside of a SQL statement (performed
in Visual Basic for Applications code for example).
This behavior is by design.
You can test for a single bit set in a long integer or integer field using
the following algorithm:
( <value> \ (2^<bit>) ) mod 2
This expression will return 1 if the <bit> is set and 0 if <bit> is not
set. <bit> is numbered from 0 to 30 inclusive where <bit>=0 is the first
bit. <bit> values greater than 30 will not work with this algorithm because
Jet uses unsigned long integer values and 2^31 is one larger than the
largest unsigned long integer value and thus will cause numeric overflow
when the division is evaluated.
Note that the \ operator and not the / operator is used. The \ operator is
used for integer division. The / operator is used for floating point
division and will cause unexpected results when used with this algorithm.
You can check <bit> 31 for a long integer using the following algorithm:
iif( <value> < 0, 1, 0 )
This works because an unsigned long integer that is less than zero means
the highest order bit (bit 31 for a long) is set.
Suppose you have a table named Test and a long integer field named
TestFlags. You can use the following SQL statements to test to see if
the bit 11 is set in the TestFlags field
SELECT * FROM Test WHERE ([TestFlags]\2^11) mod 2 = 1
or replace 2^11 with 2048 to save some query calculation time:
SELECT * FROM Test WHERE ([TestFlags]\2048) mod 2 = 1
You can use the following SQL to test for bit 31
SELECT * FROM Test WHERE iif( [TestFlags] < 0, 1 ,0 ) = 1
but this SQL statement would be a much more efficient test for bit 31:
SELECT * FROM Test WHERE [TestFlags] < 0
Note that you can also create calculated columns in SQL to display the
results of one or more bit checks:
SELECT ([TestFlags]\2^11) mod 2 AS Bit11Set FROM Test
You can run the following ADO code to verify that this algorithm works
correctly over various ranges and with various bit flags. Note the test
requires a blank Microsoft Access database named C:\Db1.mdb and a reference
to Microsoft ActiveX Data Objects.
' START SAMPLE CODE
Sub VerifyBitTest()
Dim i As Long, min As Long, max As Long, bit As Long
Dim conn As New ADODB.Connection
Dim rs As New ADODB.recordset
conn.Open "DRIVER=Microsoft Access Driver (*.mdb);" & _
"MAXBUFFERSIZE=128;DBQ=c:\db1.mdb"
On Error Resume Next
conn.Execute "DROP TABLE Test"
On Error GoTo 0
conn.Execute "CREATE TABLE Test (TestFlags LONG)"
conn.Execute "INSERT INTO Test (TestFlags) VALUES (0)"
min = 2 ^ 0: max = 2 ^ 30: bit = 11
For i = min To max ' This could take a while.
rs.Open "SELECT (" & i & "\2^" & bit & _
") mod 2 AS BitSet FROM Test", conn
If rs!BitSet <> IIf((i And (2 ^ bit)) > 0, 1, 0) Then
MsgBox "Bit Test Failure!"
Exit Sub
End If
rs.Close
DoEvents
If i Mod 100 = 0 Then Debug.Print "Verified " & i & " of " & max
Next i
End Sub
' END SAMPLE CODE