/e modifier to s///; short "1 liner" ?

Pls consider:

$ echo 'Dec 9'|perl -lpe 's/(Dec)(\s+)(\d+)/printf "$1$2 %d--huh?:",$3 + 5;/e;'
Dec 14--huh?:1

How do I get rid of the "1" in "--huh?:1". Also why is the "1" there?

If there is a cleaner approach to the above 1 liner, that would be fine
also - I just want preserve the rest of the input,
and add an integer offset to the date.

--
thanks much,
Tom

Re: /e modifier to s///; short "1 liner" ?

In article <4759904e$0$11892$afc38c87@>,
Name withheld by request wrote:
>Pls consider:
>
> $ echo 'Dec 9'|perl -lpe 's/(Dec)(\s+)(\d+)/printf "$1$2 %d--huh?:",$3 + 5;/e;'
> Dec 14--huh?:1
>
>How do I get rid of the "1" in "--huh?:1". Also why is the "1" there?

still not sure we the "1" is coming from

>If there is a cleaner approach to the above 1 liner, that would be fine
>also - I just want preserve the rest of the input,
>and add an integer offset to the date.

I was not thinking about the -p switch very clearly...
This works for me:

$ echo 'a b c Dec 9 d e f'|perl -lne 's/^(.*?)(Dec)(\s+)(\d+)(.*)$/printf "$1$2$3 %d$5\n",$4 + 5/e;'
a b c Dec 14 d e f

Re: /e modifier to s///; short "1 liner" ?

anonb6e9@nyx.net (Name withheld by request) wrote:
> In article <4759904e$0$11892$afc38c87@>,
> Name withheld by request wrote:
> >Pls consider:
> >
> > $ echo 'Dec 9'|perl -lpe 's/(Dec)(\s+)(\d+)/printf "$1$2 %d--huh?:",
> > $3 + 5;/e;'
> > Dec 14--huh?:1
> >
> >How do I get rid of the "1" in "--huh?:1". Also why is the "1" there?
>
> still not sure we the "1" is coming from

The evaluation of the printf has both a side-effect and a value.
The side effect is the printing of 'Dec 14--huh?:'. The return
value is "1", for success. The part of the thing in $_ that matched gets
replaced with this "1", because that is what s///e does. Due to the -p
switch, the value of $_ at the end of the implicit loop, namely "1", gets
printed.


> >If there is a cleaner approach to the above 1 liner, that would be fine
> >also - I just want preserve the rest of the input,
> >and add an integer offset to the date.

You should probably use sprintf. sprintf returns the formatted string,
and that return value would then get substituted into $_. printf doesn't
return the formatted string, it prints it to some filehandle and then
returns "1" upon success.

Xho

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Re: /e modifier to s///; short "1 liner" ? <Thx for soln>

--snip
>> > $ echo 'Dec 9'|perl -lpe 's/(Dec)(\s+)(\d+)/printf "$1$2 %d--huh?:",
>> > $3 + 5;/e;'
>> > Dec 14--huh?:1
>> >
>> >How do I get rid of the "1" in "--huh?:1". Also why is the "1" there?
>>
>> still not sure we the "1" is coming from
>
>The evaluation of the printf has both a side-effect and a value.
>The side effect is the printing of 'Dec 14--huh?:'. The return
>value is "1", for success. The part of the thing in $_ that matched gets
>replaced with this "1", because that is what s///e does. Due to the -p
>switch, the value of $_ at the end of the implicit loop, namely "1", gets
>printed.

Thanks for the detailed explanation.

--snip
>You should probably use sprintf. sprintf returns the formatted string,
>and that return value would then get substituted into $_. printf doesn't
>return the formatted string, it prints it to some filehandle and then
>returns "1" upon success.

Thanks again

Re: /e modifier to s///; short "1 liner" ? <test of soln, works fine>

per your suggestion:

$ echo 'a b Dec 9 c d'|perl -lpe 's/(Dec)(\s+)(\d+)/sprintf "$1$2%d",$3 + 5/e;'
a b Dec 14 c d

works just fine

Re: /e modifier to s///; short "1 liner" ?

At 2007-12-07 02:55PM, "Name withheld by request" wrote:
> >Pls consider:
> >
> > $ echo 'Dec 9'|perl -lpe 's/(Dec)(\s+)(\d+)/printf "$1$2 %d--huh?:",$3 + 5;/e;'
> > Dec 14--huh?:1
> >
> >How do I get rid of the "1" in "--huh?:1". Also why is the "1" there?
>
> >If there is a cleaner approach to the above 1 liner, that would be fine
> >also - I just want preserve the rest of the input,
> >and add an integer offset to the date.

Date calculations should be done with date modules. You probably don't
want to see "Dec 32".

echo 'Dec 9' | perl -MDate::Manip -lne '
print UnixDate(DateCalc(ParseDate($_),ParseDateDelta("+5d")), "%b %e")
'

Note I use -n instead of -p.

--
Glenn Jackman
"You can only be young once. But you can always be immature." -- Dave Barry