gmane.comp.shells.zsh.user

Re: zsh-users Digest 21 May 2012 17:55:56 -0000 Issue 1820

Ronald Fischer — 2012-05-22T07:38:56


Doesn't look like:

(1;0) /cygdrive/c> type compinit
compinit not found

However I think I should have the completion system installed:

(1;1) /cygdrive/c> zmodload
zsh/compctl
zsh/complete
zsh/main
zsh/zle

Is there an additional module that I need here?



Thank you for this suggestion. I'll have a look at it!

Ronald

Re: completion oddity

Bart Schaefer — 2012-05-22T06:35:07

}
} I did a quick test with the matcher-list zstyle, though, but couldn't
} get it to work (without the -M "" in _k).

Two things ...

}     zstyle ":completion::complete:k::" matcher-list ""

(1) The matcher-list is looked up very early in completion.  The context
isn't yet this specific at that point.  (I think this is in the FAQ.)

}     zstyle ":completion:*" matcher-list ""

The matcher-list style is only used when it is not empty!

_description: [[ -n "$_matcher" ]] && opts=($opts -M "$_matcher")

So you can't use matcher-list to turn off matching, only to modify it.

} I also took another look through the GNU tar options for a similar
} situation, and found

Hmm, when I try gtar completion I get

_arguments:comparguments:312: invalid option definition: -[0-7]lmh[specify drive and density]

So I'm not really able to reproduce this.

}     gtar --show--names<TAB>
} 
} and it beeps and moves me to between the second double-hyphen, then after
} another beep, starts cycling between the two.  If I just have

Re: completion oddity

Danek Duvall — 2012-05-22T05:16:17

I also took another look through the GNU tar options for a similar
situation, and found

    --show-transformed-names
    --show-stored-names

And tried

    gtar --show--names<TAB>

and it beeps and moves me to between the second double-hyphen, then after
another beep, starts cycling between the two.  If I just have "--show"
there, then it ends up cycling among four alternatives.  Perhaps this isn't
sufficiently similar to my test case, but I'm not sure why not.  I'll take
a closer look tomorrow, but if it's easy to explain, I wouldn't mind.

Oh, and the default match spec is documented in the -M option to
_arguments. 

Thanks,
Danek

Re: completion oddity

Danek Duvall — 2012-05-22T05:01:58


Wonderful.  I'll have to spend some time tomorrow figuring out exactly how
I want to deal with this.  While your explanation makes sense to me, I'm
not sure that it's a natural behavior, or that I could convince anyone else
of it.  I'm tempted to just put -M "" in where needed, but I kinda like the
idea of the multiple partial-word matching, even if it'd never occurred to
me to try it before myself.

I did a quick test with the matcher-list zstyle, though, but couldn't get
it to work (without the -M "" in _k).  I tried both

    zstyle ":completion::complete:k::" matcher-list ""
    zstyle ":completion:*" matcher-list ""

(the first being the string I get when executing _complete_help, and the
second just being super-inclusive) but after both, I got the original
behavior.  I also tried "matcher", but no dice.  Any help here would be
appreciated.

Thanks,
Danek

Re: completion oddity

Bart Schaefer — 2012-05-22T02:42:32

}
}     _k () { _arguments --r1-word --really-r1-word }
} 
} then the second tab simply completes --r1-word, which seems wrong to me.

It's the hyphens.

Completion works on words.  When you have a compound word joined with
hyphens, it tries to split it up and complete the individual components.
In cases where this is ambiguous, it fills in any word fragments that
are common to all the possible choices and leaves the cursor at the
first ambiguous location.  (Most of this can be changed with zstyle,
for the predefined completion set.)

With --r-word on the line, there is only one possible completion that
has exactly one embedded hyphen.  This is considered a unique match, so
when you press TAB, it fills in that word and is done.  If that isn't
what you want, it's up to you to type (in this example) one of "e" or
"-" at the point where the cursor was left, to disambiguate the choice
before attempting completion again.

Keep in mind that completion is designed to minimize keystrokes, not
to minimize the number

completion oddity

Danek Duvall — 2012-05-21T21:40:23

If I have a completion function like this:

    _k () { _arguments --r1-word --r2-word }

and I press TAB at the end of the following commandline:

    % k --r

then I get

    % k --r-word

Another tab beeps and lists the choices, a third tab beeps and chooses
--r1-word, and thereon tabs simply cycle between the two.  That's more or
less expected.

If I define _k instead as

    _k () { _arguments --r1-word --really-r1-word }

then the second tab simply completes --r1-word, which seems wrong to me.

This is all with "zsh -f", with nothing but

    autoload -Uz compinit
    compinit -i
    compdef _k k

run before attempting the completion.

A co-worker says he noticed this behavior change (from better to worse)
sometime around the switch in Solaris from 4.3.10 to 4.3.12, but I can
reproduce it on our builds that had 4.3.9, so I'm guessing it's been around
for a while.

Thanks,
Danek

Re: Extracting the 4th word of the first line in a file - is there a more elegant solution?

Bart Schaefer — 2012-05-21T19:48:47

}
} You can do it without an auxiliary process, but with a local variable,
} which is probably more efficient and the best compromise:
} 
}   read -A line <$1
}   field=$line[4]

If you're already using a local throwaway,

    read x x x field x <$1

Or you can use the omnipresent local argv so you needn't declare it:

    (){ read -A argv; field=$4 } < $1

} You might want "read -qe" if the file contains backslashes.

I believe -q is a typo for -r there.

Re: global aliases substituting within a path

Bart Schaefer — 2012-05-21T17:54:48

}
} > echo a/f/b/b/x<TAB>
} 
} Doesn't help here, for two reason:
} 
} (1) At least in my version of zsh (4.3.12), this would not expand the
} interim directories (f,b,b).

Does your zsh configuration include installing the completion system
(the "compinit" command)?

} (2) Even if I could do TAB completion on directories within the path,
} this is not what I'm looking for, because it is still cumbersome to
} type.

You might want to look at named directories:

    hash -d aaa=aaa/foo/bar/baz
    hash -d bbb=bbb/foo/bar/baz
    hash -d ccc=ccc/foo/bar/baz

Then you can write

    ls ~aaa/xxx
    ls ~bbb/yyy
    ls ~ccc/zzz

I don't know how much variation there is in the "foo/bar/baz" part of
your structure.  You can try using dynamic named directories:

    X=foo/bar/baz
    ls ~[aaa/X]/xxx

That's not saving any keystrokes over aaa/$X/xxx but might have some
aesthetic advantages.  It's implemented with something like this:

    stem_name_hook() {
case $1 in
        (n) local root=$2:h stem=$2:t

Re: Hi, I'm new comer.

zzapper — 2012-05-19T12:51:06




A good place to start

http://zshwiki.org/

Re: Extracting the 4th word of the first line in a file - is there a more elegant solution?

Peter Stephenson — 2012-05-18T13:30:25

On Fri, 18 May 2012 15:16:54 +0200
Ronald Fischer <ynnor< at >mm.st> wrote:

It depends a bit how big the file is.  If it's small, it's very easy,
and involves no extra processes (the "$(<...)" is optimised to a read):

  field=${$(<$1)[4]}

The disadvantage is that reads the entire file into memory first.  It's
hard to think of a way that avoids that that doesn't use an auxiliary
process, but apart from that it's still relatively straightforward
within zsh:

  field=${$(read -e <$1)[4]}

This reads a single line from the file and echoes it instead of
assigning it to a parameter.  You might want "read -qe" if the file
contains backslashes.

You can do it without an auxiliary process, but with a local variable,
which is probably more efficient and the best compromise:

  read -A line <$1
  field=$line[4]

Extracting the 4th word of the first line in a file - is there a more elegant solution?

Ronald Fischer — 2012-05-18T13:16:54

Within a script, I have need to store the 4th word in the first line of
a file.

Of course this is trivial to implement. However, I'm curious whether
there exists a better way to implement than what I found.

My current solution goes like this:

# $1 is the filename
line=$(head -n 1 $1)
field=${lin[(w)4]}

This is OK, but this needs an auxiliary variable 'line'. 


I could also do it like this:

field=$(head -n 1 $1|cut  -f 4 -d ' ')

No aux variable, but needs a pipe


So I wonder, whether there is a elegant "zsh" solution, which solves
this in a more elegant way?

Ronald

Re: global aliases substituting within a path

Ronald Fischer — 2012-05-18T13:08:19

Doesn't help here, for two reason:

(1) At least in my version of zsh (4.3.12), this would not expand the
interim directories (f,b,b). Instead I have to write them out explicitly
(foo/bar/baz), i.e. this is exactly what I wanted to avoid.

(2) Even if I could do TAB completion on directories within the path,
this is not what I'm looking for, because it is still cumbersome to
type.

:-(

Ronald

Re: How do I find shortest match?

Peter Stephenson — 2012-05-17T20:04:42

On Thu, 17 May 2012 15:11:09 -0400
TJ Luoma <luomat< at >gmail.com> wrote:

I think it was some years ago now.


(I should have pointed out it needs EXTENDED_GLOB for the "^".)

First it replaces all the characters in all the elements of $MATCHES
with a question mark, then orders them in the standard collating order
with (o).  As all the characters are the same, the shortest string comes
first and [1] picks it.  So $min is a single element array containing as
many ?'s as there characters in the shortest string.

The next match "simply" eliminates from MATCHES (':#') anything that
doesn't match ('^') the pattern ${~min}, where the ~ turns the ?'s into
active pattern characters.  So this matches any element of MATCHES which
has as many characters as the shortest string.  In fact, that could be
more than one, so you really need

${${MATCHES:#^${~min}}[1]}

to select the first.

Re: How do I find shortest match?

TJ Luoma — 2012-05-17T19:11:09


On Thursday, May 17, 2012 at 3:05 PM, Peter Stephenson wrote:

I have the memory and attention span of a tsetse fly, so I'll take your word for it, although I myself have no idea what this is doing. I've long accepted the fact that zsh is smarter than I am, and only hope to make the best use of it whenever possible :-) 

TjL

ps - Thanks for the reply and solutions.

Re: How do I find shortest match?

Peter Stephenson — 2012-05-17T19:05:03

On Wed, 16 May 2012 15:18:31 -0400
TJ Luoma <luomat< at >gmail.com> wrote:

  sz() { REPLY=${(l.4..0.)${#REPLY}} }
  print -l ${^MATCHES}(o+sz[1])

Note you'll need the MATCHES to be the actual path to the file since
glob qualifiers only work on real files.

There's a really nasty trick that doesn't require them to correspond to files, which you'll hate.  I think you can combine it into a single expression, but it's bad enough in two.

  local min
  min=(${${(o)${MATCHES//?/\?}}[1]})
  print -l ${MATCHES:#^${~min}}

I'm sure it's completely obvious what this is doing.  It's been posted before.

How do I find shortest match?

TJ Luoma — 2012-05-16T19:18:31

I have a folder which has a lot of txt files, and in that folder 
are a lot of duplicate files. Most of the duplicates are 
numbered like this:

10-6- Make a universal 10-6-7 Snow Leopard installer-1.txt
10-6- Make a universal 10-6-7 Snow Leopard installer-2.txt
10-6- Make a universal 10-6-7 Snow Leopard installer-3.txt
10-6- Make a universal 10-6-7 Snow Leopard installer-4.txt
10-6- Make a universal 10-6-7 Snow Leopard installer.txt

Not not all of them. For example, I might have another identical 
file named

     todo-make-snowleopardinstaller.txt

What I want to do is go through the entire folder and find all 
duplicate files (files with identical md5sum).

Then I want to keep ONLY the one with the shortest filename.

Here's what I have so far

#!/bin/zsh

DIR=/Users/luomat/Dropbox/txt/

     # to avoid 'arg list too long'
     # note that 'gmd5sum' prints the sum
     # and then two spaces, and then the filename
ALL=$(find $DIR -type f -print0 | xargs -0 gmd5sum)

     # these are all the MD5 sums which

Re: global aliases substituting within a path

Manuel Presnitz — 2012-05-16T12:40:13

Not an answer to your questions, but also quite handy sometimes:

$ ls aaa/**/xxx

Manuel.
 
 
 
Ronald Fischer hat am 16.05.2012 um 13:43 folgendes geschrieben:

Re: global aliases substituting within a path

Valodim Skywalker — 2012-05-16T12:17:38

echo a/f/b/b/x<TAB>

 - V

global aliases substituting within a path

Ronald of Steiermark — 2012-05-16T11:41:18

I have a set of directory structures like this:

aaa/foo/bar/baz/xxx
bbb/foo/bar/baz/yyy
ccc/foo/bar/baz/zzz
etc.

I'm looking for a way to make typing easier on the command line, in
order to not have to type foo/bar/baz all the time.

Of course I can achieve this by setting a shell variable in my .zshrc:

X=foo/bar/baz

Then I can do for instance

  ls aaa/$X/xxx

Now I recently learned about global aliases, which permit alias
substitution to be done within the command line, and I thought that I
maybe could use this. Here was my (failed) attempt:

alias -g X=foo/bar/baz

# Does NOT work at hoped
ls aaa/X/xxx

X is not substituted, because it is not a word on its own (not
surrounded by spaces).

My question: For my problem, do I have to stick with my original
solution (shell variable), or is it a way to do it with aliases, or is
there maybe an even more clever way to achieve my goal?

Ronald

global aliases substituting within a path

Ronald Fischer — 2012-05-16T11:43:13

I have a set of directory structures like this:

aaa/foo/bar/baz/xxx
bbb/foo/bar/baz/yyy
ccc/foo/bar/baz/zzz
etc.

I'm looking for a way to make typing easier on the command line, in
order to not have to type foo/bar/baz all the time.

Of course I can achieve this by setting a shell variable in my .zshrc:

X=foo/bar/baz

Then I can do for instance

  ls aaa/$X/xxx

Now I recently learned about global aliases, which permit alias
substitution to be done within the command line, and I thought that I
maybe could use this. Here was my (failed) attempt:

alias -g X=foo/bar/baz

# Does NOT work at hoped
ls aaa/X/xxx

X is not substituted, because it is not a word on its own (not
surrounded by spaces).

My question: For my problem, do I have to stick with my original
solution (shell variable), or is it a way to do it with aliases, or is
there maybe an even more clever way to achieve my goal?

Ronald

Re: changing case to Title Case

Peter Stephenson — 2012-05-15T19:29:59

On Mon, 14 May 2012 15:30:49 -0400
TJ Luoma <luomat< at >gmail.com> wrote:

If you like squiggles, you could do this:

print -r -- ${${(L)foo}//(#b)((#s)|[[:space:]])([[:alpha:]])/$match[1]${(U)match[2]}}

gmane.comp.shells.zsh.user

Re: zsh-users Digest 21 May 2012 17:55:56 -0000 Issue 1820

Re: completion oddity

Re: completion oddity

Re: completion oddity

Re: completion oddity

completion oddity

Re: Extracting the 4th word of the first line in a file - is there a more elegant solution?

Re: global aliases substituting *within* a path

Re: Hi, I'm new comer.

Re: Extracting the 4th word of the first line in a file - is there a more elegant solution?

Extracting the 4th word of the first line in a file - is there a more elegant solution?

Re: global aliases substituting *within* a path

Re: How do I find shortest match?

Re: How do I find shortest match?

Re: How do I find shortest match?

How do I find shortest match?

Re: global aliases substituting *within* a path

Re: global aliases substituting *within* a path

global aliases substituting *within* a path

global aliases substituting *within* a path

Re: changing case to Title Case

Re: global aliases substituting within a path

Re: global aliases substituting within a path

Re: global aliases substituting within a path

Re: global aliases substituting within a path

global aliases substituting within a path

global aliases substituting within a path