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Re: insert-last-word/copy-prev-word/... question
- X-seq: zsh-users 4737
- From: Dominik Vogt <dominik.vogt@xxxxxx>
- To: zsh-users@xxxxxxxxxx
- Subject: Re: insert-last-word/copy-prev-word/... question
- Date: Wed, 6 Mar 2002 11:57:38 +0100
- In-reply-to: <20020306112257.A16104@xxxxxxxxxxx>
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- References: <20020301153313.A8129@xxxxxxxxxxx> <20020306112257.A16104@xxxxxxxxxxx>
- Reply-to: dominik.vogt@xxxxxx
On Wed, Mar 06, 2002 at 11:22:57AM +0100, poeml@xxxxxxxxxxx wrote:
> On Fri, Mar 01, 2002 at 03:33:13 +0100, Dominik Vogt wrote:
> > Let's assume I have this in the history
> > ls xxx xxx foo
> > ls xxx xxx bar
> > ls yyy zzz baz
> > And I'm typing a new command line:
> > $ ls first second next
> > ^
> > cursor
> > with insert-last word, I can copy "baz", "bar", "foo" to the
> > cursor position. With copy-prev-word I can copy "next". But I'd
> > like to
> > - Call some function multiple times. With the first call I get
> > "next". WIth the second call I get "second" and with the
> > third call I get "first".
> > - The same should work on previous lines in the history: First I
> > call insert-last and get "baz", then I call said function and
> > get "zzz", then I call it again and get "yyy".
> > Is that possible?
> How about simply prefixing insert-last with a multiplier, is that what
> you want (not sure if I understand correctly)?
> Like "meta-2 meta-." which would give you "zzz" in you example.
Ah, yes. I guess you have to already know what the argument to
insert-last-word does to understand the man page :-) Now, is
there a way to call that with increasing argument values, i.e.
first call = Meta-1 insert-last-word
second call = Meta-2 insert-last-word
third call = Meta-3 insert-last-word
Dominik ^_^ ^_^
Dominik Vogt, email: d.vogt@xxxxxxxxxxx
LifeBits Aktiengesellschaft, Albrechtstr. 9, D-72072 Tuebingen
fon: ++49 (0) 7071/7965-0, fax: ++49 (0) 7071/7965-20
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