Is it possible to use the L processing code on a multi-valued field to
get the length of each value?

If so, how?


Thanks,
Danny

"ddspell-m3" <ddspell (AT) yahoo (DOT) com> wrote

Not sure what flavour you're using, but on D3, the following will work in
the attribute 8 of the relevant dictionary:

F{mv#}]L0

Where {mv#} should be replaced with your multi-valued attribute number and
the square bracket is actually a value mark.

HTH

With Kindest Regards

Mike Wooding

Mike's solution works just fine - and in Universe (using Mike's notation)
this also does the trick (without the value mark):
F;(mv#);(L0) - example: F;5;(L0) where attribute 5 has is the multi-valued
data.
Tom
"Mike Wooding" <mikewooding (AT) email (DOT) com> wrote

I think you can just use the "L" correlative by itself- it does not
need to be wrapped in an "F" correlative. And I don't think you need
the zero either, i.e. just put "L" in attribute 8 of a (pick type) dict
item, reference the field you want the length of in attribute 2, and
that's it. Works the same for single or multi-value fields in both D3
and Universe.

/Scott Ballinger
Pareto Corporation
Edmonds WA USA
206 713 6006

Tom Phillips wrote:

Scott Ballinger wrote:
Yes, this works for me since I'm using D3.

The only odd thing is that I cannot mix a single and multi-valued field
in my LIST statement. You see I have a field named LIS which is a
group identifier and LI.PARTS is a member. I setup the correlative as
LEN.PARTS, then when I do LIST SO LIS LEN.PARTS, I can see the results.
But if I do, LIST SO LIS LI.PARTS LEN.PARTS, the results for LEN.PARTS
are not displayed. I tried LIST SO LEN.PARTS LIS LI.PARTS, but that
didn't work either. In this case, I get the results for LEN.PARTS, but
I don't see LI.PARTS. If I do LIST SO LEN.PARTS, I don't get the
results for LEN.PARTS, so having the group identifier is necessary, but
I can't have both LI.PARTS and LEN.PARTS in the same listing.


Thanks,
Danny