"David W. Tamkin" <dattier(_at_)ripco(_dot_)com> writes:
I have this code, and it is working:
...
:0E
* cond3
* cond4
* cond5
{
:0wc:local.lock
| pipe1
:0ec # needs no local lock
| pipe2
}
If cond1 and cond2 are false, and moreover at least one among cond3, cond4,
and cond4 are false, then none of the above code is executed; that is inten-
tional.
Now, here's my question: is the following equivalent?
...
:0Ewc:local.lock
* cond3
* cond4
* cond5
| pipe1
:0Aec # needs no local lock
| pipe2
Also, what if I leave off the `A'?
Those should be equivalent. Using 'A' and 'e' together is perfectly
legal. The match/no-match status of the ':0Ewc' recipe will be used
directly by the 'A' flag, regardless of whether that recipe didn't match
because of its 'E' flag or because of a failed condition.
Philip Guenther
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