- From: pete@xxxxxxxxxxxxxxxxxx
- Date: Wed, 04 Apr 2007 16:14:38 +0000 (UTC)
In article <1175683382.210342.243490@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>
david a kennedy
does anybody no how to create a relation between 2 dbfs.
if i post the exe and the 2 dbfs up here will someone try and help.
i know i have to have the 2 dbfs open together.
all help appreciated
Please do *not* post the exe here! (a) this is a discussion
[text] group, no binaries and (b) it will serve no purpose, as
the change you need to make is to the source.
Assuming that you want to keep dbf_B record pointer at a place
that depends on the position of the record pointer in dbf_A:
1. open dbf_B
2. create (or open) its index whose key corresponds to a field or
expression based on a field in dbf_A
3. open dbf_A
4. set the relation to <expr> into <dbf_B index>
Or something like that... read the manual.
Alternatively, if the two dbfs have the same number of records
and all you want to do is to keep the record numbers in sync, you
can skip step (2) above and use RECNO() as the expression.
If that doesn't help, please post more details (but not the exe!).
"We have not inherited the earth from our ancestors,
we have borrowed it from our descendants."
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