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".
Related
I have a folder resembling this structure:
nietzsche.txt
kant.org
buddha.txt
kierkegaard.org
aristotle.txt
plato.org
I wish to read the text files that have the *.org extension, so I use the command:
ls | grep .org
The above command neatly sends the following to stdin:
> kant.org
> kierkegaard.org
> plato.org
I would like to open the files listed above in vim all at once - with the above given example, this is trivial; it would just mean typing out the list of files prefixed with "vim", for example:
vim kant.org kierkegaard.org plato.org
...but in my actual folder of articles there are several hundred plain text files, with the *.org and the *.txt extension. It isn't a matter of converting the org files to true plain text, it's trying to get vim to use the output of other commands through pipes. In reality, the conditions for generating the "books-to-read" list are far more complicated (ie. using date last read, author, date written etc) so a simple find and replace of org-to-txt wouldn't work, as I currently have a bash script to generate the list and spit it to stdout.
How would I get vim to accept the output of a command like grep as a list of files to open immediately?
In this specific example, ls | grep .org is pointless since you can simply do:
$ vim *.org
As for the general case, you would use $ man xargs on Unix-like systems:
$ <command that generates a list of files> | xargs -o vim
or:
$ <command that generates a list of files> | xargs vim --not-a-term
Note that xargs' -o and Vim's --not-a-term are more or less the opposite of each other. The former ensures that xargs passes a proper tty to Vim, while the latter ensures that Vim doesn't complain if there is no attached tty.
You can use command mode completion inside vim:
:e *.org<C-a>
Read more on :h c_CTRL-A
How can I use ls (or other commands) and grep together to search from specific files for a certain word inside that file?
Example I have a file - 201503003_315_file.txt and I have other files in my dir.
I only want to search files that have a file name that contains _315_ and inside that file, search for the word "SAMPLE".
Hope this is clear and thanks in advance for any help.
You can do:
ls * _315_* | xargs grep "SAMPLE"
The first part: ls * _315_* will list only files that have 315 as part of the file name, this list of files is piped to grep which will scan each one of them and look for "SAMPLE"
UPDATE
A bit easier (and actually safer) approach was mentioned by David in the comments bellow:
grep "SAMPLE" *_315_*.txt
The reason why it's safer is that ls doesn't handle well special characters.
Another option, as mentioned by Charles Duffy in the comments below:
printf '%s\0' *_315_* | xargs -0 grep
Change to that directory (using cd dir) and try:
grep SAMPLE *_315_*
If you really MUST use ls AND grep try this:
ls *_315_* | xargs grep SAMPLE
The first example, however, requires less typing...
This question already has answers here:
How to pass command output as multiple arguments to another command
(5 answers)
Read expression for grep from standard input
(1 answer)
Closed last month.
I am looking for insight as to how pipes can be used to pass standard output as the arguments for other commands.
For example, consider this case:
ls | grep Hello
The structure of grep follows the pattern: grep SearchTerm PathOfFileToBeSearched. In the case I have illustrated, the word Hello is taken as the SearchTerm and the result of ls is used as the file to be searched. But what if I want to switch it around? What if I want the standard output of ls to be the SearchTerm, with the argument following grep being PathOfFileToBeSearched? In a general sense, I want to have control over which argument the pipe fills with the standard output of the previous command. Is this possible, or does it depend on how the script for the command (e.g., grep) was written?
Thank you so much for your help!
grep itself will be built such that if you've not specified a file name, it will open stdin (and thus get the output of ls). There's no real generic mechanism here - merely convention.
If you want the output of ls to be the search term, you can do this via the shell. Make use of a subshell and substitution thus:
$ grep $(ls) filename.txt
In this scenario ls is run in a subshell, and its stdout is captured and inserted in the command line as an argument for grep. Note that if the ls output contains spaces, this will cause confusion for grep.
There are basically two options for this: shell command substitution and xargs. Brian Agnew has just written about the former. xargs is a utility which takes its stdin and turns it into arguments of a command to execute. So you could run
ls | xargs -n1 -J % grep -- % PathOfFileToBeSearched
and it would, for each file output by ls, run grep -e filename PathOfFileToBeSearched to grep for the filename output by ls within the other file you specify. This is an unusual xargs invocation; usually it's used to add one or more arguments at the end of a command, while here it should add exactly one argument in a specific place, so I've used -n and -J arguments to arrange that. The more common usage would be something like
ls | xargs grep -- term
to search all of the files output by ls for term. Although of course if you just want files in the current directory, you can this more simply without a pipeline:
grep -- term *
and likewise in your reversed arrangement,
for filename in *; do
grep -- "$#" PathOfFileToBeSearched
done
There's one important xargs caveat: whitespace characters in the filenames generated by ls won't be handled too well. To do that, provided you have GNU utilities, you can use find instead.
find . -mindepth 1 -maxdepth 1 -print0 | xargs -0 -n1 -J % grep -- % PathOfFileToBeSearched
to use NUL characters to separate filenames instead of whitespace
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