> So I guess I'm saying that in the code if you change the $asinteger
> assignment to:
> $asinteger = "NULL"; /* notice the quotes */
> it may work.  This would make the $asinteger PHP variable a string, but
> since the SQL would either like to have a numeric value there or the
> keyword NULL, that is probably what you want.

It doesn't work. I think that I need to modify interbase.c in php source
code. See Jeremy Bettis's code below!

Thanx for your answer,


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