On Jan 8, 2008 9:24 AM, Robert Citek <[EMAIL PROTECTED]> wrote:
> On Jan 8, 2008 8:59 AM, Chas. Owens <[EMAIL PROTECTED]> wrote:
> > On Jan 7, 2008 6:59 PM, Robert Citek <[EMAIL PROTECTED]> wrote:
> > > How can I get rid of the @foo?
> > >
> > > This does what I want, but uses the "temporary" variable @foo:
> > >
> > > $ ls | perl -e '@foo=<>; chomp @foo ; print join(" ", @foo)'
> > >
> > > This eliminates the temporary variable but doesn't work:
> > >
> > > $ ls | perl -e 'print join(" ", chomp(<>))'
> > >
> > > Thanks in advance.
> > >
> > > Regards,
> > > - Robert
> >
> > It looks like you are trying to turn this
> >
> > foo
> > bar
> > baz
> >
> > into this
> >
> > foo bar baz
> >
> > A much better way is this
> >
> > perl -pe 's/\n/ /';
>
> That would be a better perl way if I was just trying to concatenate a
> list of items. If I was, an even better way would be to substitue
> xargs for perl:
>
> $ ls | xargs
>
> What I really want to know is if there's a way to eliminate the
> temporary variable. If not, that's fine. I was just wondering if
> there's something I'm overlooking. Come to think of it, maybe I could
> use map.
Here's an example closer to the real problem:
$ ls | perl -e '@F=(<>) ; chomp @F ; print "('\''" .
join("'\'',\n'\''",@F) . "'\'') \n" ; '
Regards,
- Robert
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