I would like to list all plain files that are not python scripts in zsh.
Why does the following "code" not work and what is the proper solution?
ls -l *(.)~*.py
UPDATE:
I have setopt extended_glob in my .zshrc.
And
ls -ld *~*.py``
works as expected.
(I added the -d in the command to prevent directories from getting expanded).
The problem is that ~ is a glob operator (that also requires EXTENDED_GLOB be set), while (.) is a glob qualifier, which means it must be added to the end of the entire pattern, not used in the middle. Use
ls *~*.py(.)
instead. That is, *~*.py is your pattern (all files not ending in .py), and (.) is applied to the results. (Perhaps yet another way to put it is to say that glob operators can only work on unqualified patterns.)
Related
Trying to find all files except those inside vendor/ folders, but why is this failing?
setopt extendedglob
for file in **/*~vendor/; do
done
See if this does what you're looking for:
setopt extendedglob
print -l ^vendor/**/*(.)
The *~ negation syntax usually needs parentheses in order to determine where the expression after the tilde ends. Your pattern is requesting all files and folders except those where the glob result name ends with vendor/. The glob result never includes the trailing slash, so you end up with all of the files and folders.
Adding parens will change the behavior of that pattern, but probably not in a useful way. This will result in a list of all of the directories where the last component is not vendor:
print -l **/(*~vendor)/
so x/y, x/y/vendor/z, and vendor/a will be included, but x/y/vendor will not.
The parentheses limit the 'not' pattern to just one piece of the path. In order to exclude matches at the top-level, the tested component needs to be at the front of the pattern:
print -l (*~vendor)/**/*
The very first pattern above uses the ^ syntax to produce the same results. The (.) glob qualifier in that pattern limits the globbing to plain files, so directories are not included.
Another variation that may be useful - this will exclude directories that have any component named vendor. It is similar to find -prune:
print -l (^vendor/)#*(.)
This will produce a list of all files except those in subdirectories with names like vendor/x, a/vendor and a/vendor/b.
EDIT #1 : I'm under the limit that all arguments are enclosed in two quotes, so that shell do not expand any argument with * to the corresponding path.
EDIT #2 : In order to retrieve directories such as */*, ../*, and dirA/*/file.out, How should I use iteration loop or recursive call?
I have just learned about the function fnmatch(). But I don't know start place.
There are many possible cases. I'm confused dealing with these all cases.
For example, Let me assume that executable program is a.out.
$./a.out -l */*
$./a.out -l ../*
$./a.out -l [file_name] [directory_name]
/* Since I also have to implement ls command with no wildcard. */
What should I do? Any advice would be awesome.
Thank you in advance.
Your problem is : shell replaces wildcard caracter * with all of the filenames matching the pattern.
Solution:
If you do not want to use this feature of bash, just put quotation marks around your command line arguments.
Calling your program that way will have the original arguments, containing wildcards.
After this, you can list all the filenames with their paths. For example using some recursive algorithm. Then you can apply some matching to these path string. (when visiting it)
If you want to be a good unix citizen, the rule is Don't do filename globbing unless you are writing a shell.
You want to write an ls-like program? Don't do any wildcard expansion. Don't treat "*" specially. Just treat your argv as a list of filenames. If your program handles these cases:
./a.out file1
./a.out file1 file2 file3
Then it will also handle
./a.out file*
correctly because the shell will do the expansion and your program won't need to know about it. And besides that, it will handle this:
zsh% ./a.out **/file<40-185>~file<90-100>(.mm-30OL[1,2])
which in zsh expanded glob syntax means: expand file40 through file185, except for file90 through file100, include only the ones that have been modified in the last 30 minutes, and use only the largest 2 files in the resulting set.
fnmatch is never going to do anything like that. But these fancy globs can be used with any command that just takes a filename list and doesn't care where it came from.
When you're in a situation where you can't take a list of filenames from the command line, then consider using fnmatch. ls isn't one of those situations.
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)
Is there a way I can get the pwd in an alias in my .zshrc file? I'm trying to do something like the following:
alias cleanup="rm -Rf `pwd`/{foo,bar,baz}"
This worked fine in bash; pwd is always the directory I've cd'd into, however in zsh it seems that it's evaluated when the .zshrc file is first loaded and always stays as my home directory. I've tested using with a really simple alias setup, but it never changes.
How can I have this change, so that calling the alias from a subdirectory always evaluates as that subdir?
EDIT: not sure if this will help, but I'm using zsh via oh-my-zsh on the mac.
When your .zshrc is loaded, the alias command is evaluated. The command consists of two words: a command name (the builtin alias), and one argument, which is the result of expanding cleanup="rm -Rf `pwd`/{foo,bar,baz}". Since backquotes are interpolated between double quotes, this argument expands to cleanup=rm -Rf /home/unpluggd/{foo,bar,baz} (that's a single shell word) where /home/unpluggd is the current directory at that time.
If you want to avoid interpolation at the time the command is defined, use single quotes instead. This is almost always what you want for aliases.
alias cleanup='rm -Rf `pwd`/{foo,bar,baz}'
However this is needlessly complicated. You don't need `pwd/` in front of file names! Just write
alias cleanup='rm -Rf -- {foo,bar,baz}'
(the -- is needed if foo might begin with a -, to avoid its being parsed as an option to rm), which can be simplified since the braces are no longer needed:
alias cleanup='rm -Rf -- foo bar baz'
In a Linux or Mac environment, Vim’s glob() function doesn’t match dot files such as .vimrc or .hiddenfile. Is there a way to get it to match all files including hidden ones?
The command I’m using:
let s:BackupFiles = glob("~/.vimbackup/*")
I’ve even tried setting the mysterious {flag} parameter to 1, and yet it still doesn’t return the hidden files.
Update: Thanks ib! Here’s the result of what I’ve been working on: delete-old-backups.vim.
That is due to how the glob() function works: A single-star pattern
does not match hidden files by design. In most shells, the default
globbing style can be changed to do so (e.g., via shopt -s dotglob
in Bash), but it is not possible in Vim, unfortunately.
However, one has several possibilities to solve the problem still.
First and most obvious is to glob hidden and not hidden files
separately and then concatenate the results:
:let backupfiles = glob(&backupdir..'/*').."\n"..glob(&backupdir..'/.[^.]*')
(Be careful not to fetch the . and .. entries along with hidden files.)
Another, perhaps more convenient but less portable way is to use
the backtick expansion within the glob() call:
:let backupfiles = glob('`find '..&backupdir..' -maxdepth 1 -type f`')
This forces Vim to execute the command inside backticks to obtain
the list of files. The find shell command lists all files (-type f)
including the hidden ones, in the specified directory (-maxdepth 1
forbids recursion).