How do I make ls honour .hidden files? - ls

The .hidden file is quite convenient to hide directories, what applications use and generated themselves.
How do I treat items listed in the .hidden file hidden, while using ls?

By default, files in a .hidden file folder would not be hidden themselves: check it with ls -l0.
You can hide them individually
chflags hidden .hidden/foo
chflags hidden .hidden/foo
Check also their Extended Attributes with xattr to see if they have any special attribute compared to regular files outside .hidden

GNU ls has option --hide=, so if you have that implementation, it's possible to create a simple wrapper:
#!/usr/bin/env sh
hide=
if [ -f .hidden ]; then
while IFS= read -r line; do
hide="$hide --hide=$line" # use `--ignore` to have them hidden even when `ls -a`
done < .hidden
fi
ls $hide "$#"
It's a rough example, there is a bit of edge cases to handle (like I did in ls.hidden).

Related

How to write a makefile executing make one directory level up

Can I write a wrapper makefile that will cd one level up and execute there make with all the command options I have given the wrapper?
In more detail:
Directory project contains a real Makefile with some different targets.
Directory project/resources contains the wrapper Makefile which should call Makefile in project.
When I am in my shell in directory project/resources, I execute
make TARGET
and the Makefile there just cds one directory up and calls
make TARGET
in the directory project.
Is this possible? And how?
You could use a very simple Makefile for all your sub-directories:
%:
$(MAKE) -C .. $#
% is a last resort match-anything pattern rule that will match any target... for which there is no implicit rule (GNU make has an incredibly large number of implicit rules). So, if none of your targets are covered by an implicit rule, this should work. Else you will have to tell make not to use the implicit rules it knows. This can be done (with GNU make) by calling make with the -r option:
cd project/resources
make -r <anything>
will call make in project for target <anything>. The main drawback is that the -r flag is passed to the sub-make and so the implicit rules will not apply neither in project, which can be a problem. If it is you can obtain the same effect by adding an empty .SUFFIXES target to theMakefile in project/resources:
.SUFFIXES:
%:
$(MAKE) -C .. $#
With my version of GNU make (3.82) it works like a charm and the sub-make has all the default implicit rules.
Yes, you can have a makefile which works for "any" target.
The GNU make manual discusses this in the Overriding Part of Another Makefile section:
Sometimes it is useful to have a makefile that is mostly just like another makefile. You can often use the ‘include’ directive to include one in the other, and add more targets or variable definitions. However, it is invalid for two makefiles to give different recipes for the same target. But there is another way.
In the containing makefile (the one that wants to include the other), you can use a match-anything pattern rule to say that to remake any target that cannot be made from the information in the containing makefile, make should look in another makefile. See Pattern Rules, for more information on pattern rules.
For example, if you have a makefile called Makefile that says how to make the target ‘foo’ (and other targets), you can write a makefile called GNUmakefile that contains:
foo:
frobnicate > foo
%: force
#$(MAKE) -f Makefile $#
force: ;
If you say ‘make foo’, make will find GNUmakefile, read it, and see that to make foo, it needs to run the recipe ‘frobnicate > foo’. If you say ‘make bar’, make will find no way to make bar in GNUmakefile, so it will use the recipe from the pattern rule: ‘make -f Makefile bar’. If Makefile provides a rule for updating bar, make will apply the rule. And likewise for any other target that GNUmakefile does not say how to make.
The way this works is that the pattern rule has a pattern of just ‘%’, so it matches any target whatever. The rule specifies a prerequisite force, to guarantee that the recipe will be run even if the target file already exists. We give the force target an empty recipe to prevent make from searching for an implicit rule to build it—otherwise it would apply the same match-anything rule to force itself and create a prerequisite loop!
One option: use a wrapper file to execute the commands to do that. Just be sure your target make files don't include the child directory that has the wrapper, or else you can create an endless loop. For example,
clean:
pushd .. && make clean && popd
Using the comment of user Renaud Pacalet and the answer to a different question the following one-liner is as close as I could get. The whole Makefile reads:
IGNORE := $(shell $(MAKE) -C .. $(MAKECMDGOALS))
This solutions comes with a few caveats:
Command line option -B does not get passed through to the subsequent make call.
The output of the subsequently called make process (in the project directory) is not printed to stdout.
The wrapper make process reports for any given target at the end :
make: *** No rule to make target TARGET. Stop.

How to make zsh search configuration in $XDG_CONFIG_HOME

Looking to make my ~ a cleaner place, I would like to move as much user configuration files into $XDG_CONFIG_HOME, which is ~/.config by default. So I would like to store all my zsh user files in $XDG_CONFIG_HOME/zsh/. So far already have this:
% ls $XDG_CONFIG_HOME/zsh/
histfile zsh_cache zshrc
Easy, you just have to fill your ~/.zshrc. Now the trickiest part seems to make zsh read directly $XDG_CONFIG_HOME/zsh/zshrc without sourcing it from ~/.zshrc. How would you proceed?
One may edit /etc/zsh/zshenv to set $XDG_CONFIG_HOME directories and $ZDOTDIR. This require write privilegies on this files though.
So provided that $HOME is defined when zsh read it (I don't know if it's the case), you may add to your /etc/zsh/zshenv:
if [[ -z "$XDG_CONFIG_HOME" ]]
then
export XDG_CONFIG_HOME="$HOME/.config/"
fi
if [[ -d "$XDG_CONFIG_HOME/zsh" ]]
then
export ZDOTDIR="$XDG_CONFIG_HOME/zsh/"
fi
It is good practice to not put a / at the end of any variable holding a certain path.
For example, $XDG_CONFIG_HOME/zsh translates to "$HOME/.config//zsh" and the / repeats because XDG_CONFIG_HOME ends with a /.
So I think your answer should be -
if [[ -z "$XDG_CONFIG_HOME" ]]
then
export XDG_CONFIG_HOME="$HOME/.config"
fi
if [[ -d "$XDG_CONFIG_HOME/zsh" ]]
then
export ZDOTDIR="$XDG_CONFIG_HOME/zsh"
fi
Variation to psychoslave's answer which uses ${HOME}/.zshenv to initiate the environment. No root access needed.
export XDG_CONFIG_HOME=${XDG_CONFIG_HOME:=${HOME}/.config}
export ZDOTDIR=${ZDOTDIR:=${XDG_CONFIG_HOME}/zsh}
source $ZDOTDIR/.zshenv
This was discussed on this thread on the zsh-users mailing list.
You may want to consider saving history in XDG_DATA_HOME. Specifications can be found at XDG Base Directory Specification.
Write a wrapper for zsh that executes zsh after setting the environment variable ZDOTDIR to where you want zsh to look for the config files.
See: http://zsh.sourceforge.net/Intro/intro_3.html

ZSH auto_vim (like auto_cd)

zsh has a feature (auto_cd) where just typing the directory name will automatically go to (cd) that directory. I'm curious if there would be a way to configure zsh to do something similar with file names, automatically open files with vim if I type only a file name?
There are three possibilities I can think of. First is suffix aliases which may automatically translate
% *.ps
to
% screen -d -m okular *.ps
after you do
alias -s ps='screen -d -m okular'
. But you need to define this alias for every file suffix. It is also processed before most expansions so if
% *.p?
matches same files as *.ps it won’t open anything.
Second is command_not_found handler:
function command_not_found_handler()
{
emulate -L zsh
for file in $# ; do test -e $file && xdg-open $file:A ; done
}
. But this does not work for absolute or relative paths, only for something that does not contain forward slashes.
Third is a hack overriding accept-line widget:
function xdg-open()
{
emulate -L zsh
for arg in $# ; do
command xdg-open $arg
endfor
}
function _-accept-line()
{
emulate -L zsh
FILE="${(z)BUFFER[1]}"
whence $FILE &>/dev/null || BUFFER="xdg-open $BUFFER"
zle .accept-line
}
zle -N accept-line _-accept-line
. The above alters the history (I can show how to avoid this) and is rather hackish. Good it does not disable suffix aliases (whence '*.ps' returns the value of the alias), I used to think it does. It does disable autocd though. I can avoid this (just || test -d $FILE after whence test), but who knows how many other things are getting corrupt as well. If you are fine with the first and second solutions better to use them.
I guess you can use "fasd_cd" which has an alias v which uses viminfo file to identifi files which you have opened at least once. In my environment it works like a charm.
Fast cd has other amazing stuff you will love!
Don't forget to set this alias on vim to open the last edited file:
alias lvim="vim -c \"normal '0\""

How to edit path variable in ZSH

In my .bash_profile I have the following lines:
PATHDIRS="
/usr/local/mysql/bin
/usr/local/share/python
/opt/local/bin
/opt/local/sbin
$HOME/bin"
for dir in $PATHDIRS
do
if [ -d $dir ]; then
export PATH=$PATH:$dir
fi
done
However I tried copying this to my .zshrc, and the $PATH is not being set.
First I put echo statements inside the "if directory exists" function and I found that the if statement was evaluating to false, even for directories that clearly existed.
Then I removed the directory-exists check, and the $PATH was being set incorrectly like this:
/usr/bin:/bin:/usr/sbin:/sbin:
/usr/local/bin
/opt/local/bin
/opt/local/sbin
/Volumes/Xshare/kburke/bin
/usr/local/Cellar/ruby/1.9.2-p290/bin
/Users/kevin/.gem/ruby/1.8/bin
/Users/kevin/bin
None of the programs in the bottom directories were being found or executed.
What am I doing wrong?
Unlike other shells, zsh does not perform word splitting or globbing after variable substitution. Thus $PATHDIRS expands to a single string containing exactly the value of the variable, and not to a list of strings containing each separate whitespace-delimited piece of the value.
Using an array is the best way to express this (not only in zsh, but also in ksh and bash).
pathdirs=(
/usr/local/mysql/bin
…
~/bin
)
for dir in $pathdirs; do
if [ -d $dir ]; then
path+=$dir
fi
done
Since you probably aren't going to refer to pathdirs later, you might as well write it inline:
for dir in \
/usr/local/mysql/bin \
… \
~/bin
; do
if [[ -d $dir ]]; then path+=$dir; fi
done
There's even a shorter way to express this: add all the directories you like to the path array, then select the ones that exist.
path+=/usr/local/mysql/bin
…
path=($^path(N))
The N glob qualifier selects only the matches that exist. Add the -/ to the qualifier list (i.e. (-/N) or (N-/)) if you're worried that one of the elements may be something other than a directory or a symbolic link to one (e.g. a broken symlink). The ^ parameter expansion flag ensures that the glob qualifier applies to each array element separately.
You can also use the N qualifier to add an element only if it exists. Note that you need globbing to happen, so path+=/usr/local/mysql/bin(N) wouldn't work.
path+=(/usr/local/bin/mysql/bin(N-/))
You can put
setopt shwordsplit
in your .zshrc. Then zsh will perform world splitting like all Bourne shells do. That the default appears to be noshwordsplit is a misfeature that causes many a head scratching. I'd be surprised if it wasn't a FAQ. Lets see... yup:
http://zsh.sourceforge.net/FAQ/zshfaq03.html#l18
3.1: Why does $var where var="foo bar" not do what I expect?
Still not sure what the problem was (maybe newlines in $PATHDIRS)? but changing to zsh array syntax fixed it:
PATHDIRS=(
/usr/local/mysql/bin
/usr/local/share/python
/usr/local/scala/scala-2.8.0.final/bin
/opt/local/Library/Frameworks/Python.framework/Versions/2.6/bin
/opt/local/Library/Frameworks/Python.framework/Versions/2.7/bin
/opt/local/etc
/opt/local/bin
/opt/local/sbin
$HOME/.gem/ruby/1.8/bin
$HOME/bin)
and
path=($path $dir)

Hiding the removal of intermediate files using Make

I use intermediate files in my Makefile, however make prints out the rm command that it uses to delete them all afterwards. How do I hide this print statement?
The make manual says that targets marked .SECONDARY will behave as .INTERMEDIATE but won't be automatically deleted. You could mark all the intermediate targets as secondary, and then remove the files yourself, something like
OBJECTS=foo.o bar.o
all:foo bar
#rm -f $(OBJECTS)
.SECONDARY: $(OBJECTS)
should do.

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