If you run arc cover it will show you who last edited the lines you have changed, which is good for deciding who should be reviewers.
Is there a way to have Arcanist automatically do this and then pre-populate the Reviewers: line in the commit message with those people?
You can get a delimited list of reviewers with this one liner:
arc cover | grep '^[^ ]' | paste -s -d, -
And you can specify reviewers when creating a diff, using the --reviewers option.
Therefore you should be able to do something like this:
arc diff --reviewers "$(arc cover | grep '^[^ ]' | paste -s -d, -)"
But this only works if you use Phabricator usernames as git author names.
Related
I have a directory full of files with names such as:
file_name_is_001
file_name_001
file_name_is_002
file_name_002
file_name_is_003
file_name_003
I want to copy only the files that don't contain 'is'. I'm not sure how to do this. I have tried to search for it, but can't seem to google the right phrase to find the results.
Details depend on operating system, shell, etc.
For a unix system a quite verbose but easy to understand approach could look like this (please mind that I didn't test it):
mkdir some_temporary_directory
mv *_is_* some_temporary_directory
cp * where_ever_you_want_to_copy_it
mv some_temporary_directory/* .
rmdir some_temporary_directory
You can do this using bash. First, here's a command to get you a list of files that don't contain the text _is_:
ls | grep -v "_is_"
This takes the output of ls and matches all values with DO NOT contain _is_ using grep -v.
In order to then copy these files, we need to turn the lines output by grep into arguments of cp. We can do this using xargs:
ls | grep -v "_is_" | xargs -J % cp % new_folder
From the xargs man page, it is a tool to "build and execute command lines from standard input".
real beginner in Unix commands so not sure if the following is actually possible but here goes.
Is it possible to highlight just one item in a ls output?
I.e.: in a directory I use the following
ls -l --color=auto
this lists 4 items in green
file1.xls
file2.xls
file3.xls
file4.xls
But I want to highlight a specific item, in this case file2.
Is this possible?
The ls program will not do this for you. But you could filter the results from ls through a custom script which modifies the text to highlight just one item. It would be simpler if no color was originally given; then you could match on the given filename (for example as the pattern in an awk script, or in a sed script) and modify just that one item, adding colors.
That is, certainly it is possible. Writing a sample script is a different question.
How you approach the problem depends on what you want from the output. If that is (literally) the output from ls with a single filename in color, then a script would be the normal approach. You could use grep as suggested in the other answer, which raises a few issues:
commenting on ls -l --color=auto makes it sound as if you are using GNU ls, hence likely using Linux. An appropriate tag for the question would be linux rather than unix. If you ask for unix, the answers should differ.
supposing that you are using Linux. Then likely you have GNU grep, which can do colors. That would let you do something like this:
ls -l | grep --color=always file2 |less -R
however, there is a known bug in GNU grep's use of color (see xterm FAQ "grep --color" does not show the right output).
using grep like this shows only the matching lines. For ls that might be a good choice. For matches in a manual page -- definitely not.
Alternatively, less (which is found more often on Unix systems than GNU grep) also can highlight matches (not in color) and would show the file you are looking for in context. You could do this:
ls -l | less -p file2
(Both grep and less use patterns aka regular expressions, but I left the example simple — read the documentation to learn more).
If you're a beginner I would strongly suggest you learn the grep command if you want to filter results - A Unix users best friend (mine anyway)
Use grep to only display the list items you want to see...
ls- l | grep "file2"
NOTE: This is no different to typing ls -l file2 by the way but your pattern could be expanded based on what you actually want displayed on the screen.
So if you had a directory full of files ".txt", ".xls", ".doc" and you wanted to only see ".doc" with the word "work" in the name (work1.doc) you could write:
ls -ls | grep "work" | grep "txt"
This would list work1.txt, work2.txt, work3.txt and so on.
This is a very basic example but I use grep extensively whilst in the unix shell and would advise using this to filter all results instead of colours.
A little side note using grep -v will show you everything but the pattern you give it
ls -l | grep -v ".txt" will show everything BUT .txt files.
I want to use awk to get the second directory pattern I am able to do the same using cut -d command and need to use reverse as there is a problem that space between the two dir does not remains the same.
Here is the string I just want the third pattern one marked in bold. I am able to achieve this with cut -d and rev but want to achieve it with awk.
* example_view /nas/viewstore/admin/example_view.vws
please note the space between the two string varies so we cannot use a fixed value.
I used the following command to do so please take in to consideration of the * which comes when the view is set somewhere.
cleartool lsview -cview |rev | cut -d ' ' -f 1 | rev|xargs
Use awk as shown below:
cleartool lsview -cview | awk '{print $NF}'
$NF refers to the last field and, by default, awk uses whitespace as a delimiter.
As an alternative to dogbane's answer (upvoted), to get the view storage path (for dynamic view or snapshot view), I usually use string substitution when I am in a bash script:
cd /views/myStartedView
storagePath=$(cleartool lsview -cview)
storagePath="${storagePath#*/}/"
I think "cleartool pwv" (print working view) does the job.
I'm very new to Unix, and currently taking a class learning the basics of the system and its commands.
I'm looking for a single command line to list off all of the user home directories in alphabetical order from the /etc/passwd directory. This applies only to the home directories, and not the contents within them. There should be no duplicate entries. I've tried many permutations of commands such as the following:
sort -d | find /etc/passwd /home/* -type -d | uniq | less
I've tried using -path, -name, removing -type, using -prune, and changing the search pattern to things like /home/*/$, but haven't gotten good results once. At best I can get a list of my own directory (complete with every directory inside it, which is bad), and the directories of the other students on the server (without the contained directories, which is good). I just can't get it to display the /home/user directories and nothing else for my own account.
Many thanks in advance.
/etc/passwd is a file. the home directory is usually at field/column 6, where ":" is the delimiter. When you are dealing with file structure that has distinct characters as delimiters, you should use a tool that can break your data down into smaller chunks for easier manipulation using fields and field delimiters. awk/cut etc, even using the shell with IFS variable set can do the job. eg
awk -F":" '{print $6}' /etc/passwd | sort
cut -d":" -f6 /etc/passwd |sort
using the shell to read the file
while IFS=":" read -r a b c d e home_dir g
do
echo $home_dir
done < /etc/passwd | sort
I think the tools you want are grep, tr and awk. Grep will give you lines from the file that actually contain home directories. tr will let you break up the delimiter into spaces, which makes each line easier to parse.
Awk is just one program that would help you display the results that you want.
Good luck :)
Another hint, try ls --color=auto /etc, passwd isn't the kind of file that you think it is. Directories show up in blue.
In Unix, find is a command for finding files under one or more directories. I think you are looking for a command for finding lines within a file that match a pattern? Look into the command grep.
sed 's|\(.[^:]*\):\(.[^:]*\):\(.*\):\(.[^:]*\):\(.[^:]*\)|\4|' /etc/passwd|sort
I think all this processing could be avoided. There is a utility to list directory contents.
ls -1 /home
If you'd like the order of the sorting reversed
ls -1r /home
Granted, this list out the name of just that directory name and doesn't include the '/home/', but that can be added back easily enough if desired with something like this
ls -1 /home | (while read line; do echo "/home/"$line; done)
I used something like :
ls -l -d $(cut -d':' -f6 /etc/passwd) 2>/dev/null | sort -u
The only thing I didn't do is to sort alphabetically, didn't figured that yet
I have two directories with the same list of files. I need to compare all the files present in both the directories using the diff command. Is there a simple command line option to do it, or do I have to write a shell script to get the file listing and then iterate through them?
You can use the diff command for that:
diff -bur folder1/ folder2/
This will output a recursive diff that ignore spaces, with a unified context:
b flag means ignoring whitespace
u flag means a unified context (3 lines before and after)
r flag means recursive
If you are only interested to see the files that differ, you may use:
diff -qr dir_one dir_two | sort
Option "q" will only show the files that differ but not the content that differ, and "sort" will arrange the output alphabetically.
Diff has an option -r which is meant to do just that.
diff -r dir1 dir2
diff can not only compare two files, it can, by using the -r option, walk entire directory trees, recursively checking differences between subdirectories and files that occur at comparable points in each tree.
$ man diff
...
-r --recursive
Recursively compare any subdirectories found.
...
Another nice option is the über-diff-tool diffoscope:
$ diffoscope a b
It can also emit diffs as JSON, html, markdown, ...
If you specifically don't want to compare contents of files and only check which one are not present in both of the directories, you can compare lists of files, generated by another command.
diff <(find DIR1 -printf '%P\n' | sort) <(find DIR2 -printf '%P\n' | sort) | grep '^[<>]'
-printf '%P\n' tells find to not prefix output paths with the root directory.
I've also added sort to make sure the order of files will be the same in both calls of find.
The grep at the end removes information about identical input lines.
If it's GNU diff then you should just be able to point it at the two directories and use the -r option.
Otherwise, try using
for i in $(\ls -d ./dir1/*); do diff ${i} dir2; done
N.B. As pointed out by Dennis in the comments section, you don't actually need to do the command substitution on the ls. I've been doing this for so long that I'm pretty much doing this on autopilot and substituting the command I need to get my list of files for comparison.
Also I forgot to add that I do '\ls' to temporarily disable my alias of ls to GNU ls so that I lose the colour formatting info from the listing returned by GNU ls.
When working with git/svn or multiple git/svn instances on disk this has been one of the most useful things for me over the past 5-10 years, that somebody might find useful:
diff -burN /path/to/directory1 /path/to/directory2 | grep +++
or:
git diff /path/to/directory1 | grep +++
It gives you a snapshot of the different files that were touched without having to "less" or "more" the output. Then you just diff on the individual files.
In practice the question often arises together with some constraints. In that case following solution template may come in handy.
cd dir1
find . \( -name '*.txt' -o -iname '*.md' \) | xargs -i diff -u '{}' 'dir2/{}'
Here is a script to show differences between files in two folders. It works recursively. Change dir1 and dir2.
(search() { for i in $1/*; do [ -f "$i" ] && (diff "$1/${i##*/}" "$2/${i##*/}" || echo "files: $1/${i##*/} $2/${i##*/}"); [ -d "$i" ] && search "$1/${i##*/}" "$2/${i##*/}"; done }; search "dir1" "dir2" )
Try this:
diff -rq /path/to/folder1 /path/to/folder2