I am trying to print only specific lines from the output based on keywords using grep
ls -RlA | grep foo | sed -n '1 p'
ls -RlA | grep bar | sed -n '1 p'
ls -RlA | grep foo_file
ls -RlA | grep bar_file
Is there a way to simplify these statements into just one command?
P.S: Order does not matter
Find all files or directories with the given names:
find . '(' -name foo -o -name bar -o -name foo_file -o -name bar_file ')' -ls
A more compact version using a regex:
find . -regex '.*/\(foo\|bar\|foo_file\|bar_file)' -ls
Same as above, but check that foo_file and bar_file are files, not directories:
find . '(' -name foo -o -name bar -o -name foo_file -type f -o -name bar_file -type f ')' -ls
Here in one command.
ls -la **/{foo,bar,foo_file,bar_file}
You can also use * inside {}, such as {*.txt,foo_*.zip}.
Note that it will not work if one of the field inside {} is not found.
Related
I have crafted this sed command which looks to be working fine, only it's being applied to all the files in my directory :
find . -type f -name '*.js' -not -path './node_modules/*' -exec sed -i .bak -E '
1i\
const env = require('\''env-var'\'');
s/(^|[^[:alnum:]_])process\.env\.([[:alnum:]_]+) \|\| ([[:alnum:]_]+)($|[^[:alnum:]_])/\1env.get('\''\2'\'').default('\''\3'\'')\4/g
s/(^|[^[:alnum:]_])process\.env\.([[:alnum:]_]+)($|[^[:alnum:]_])/\1env.get('\''\2'\'')\3/g
' {} \;
I wish to apply those transformations only to the files which match this grep command :
grep -r "process\.env\." --exclude-dir=node_modules
I tried using the pipe but I can't make the two working together. What's the right way to handle it?
EDIT: I tried this
➜ app-service git:(chore/adding-env-example) ✗ grep -r "process\.env\." --exclude-dir=node_modules | sed -i .bak -E '
1i\
const env = require('\''env-var'\'');
s/(^|[^[:alnum:]_])process\.env\.([[:alnum:]_]+) \|\| ([[:alnum:]_]+)($|[^[:alnum:]_])/\1env.get('\''\2'\'').default('\''\3'\'')\4/g
s/(^|[^[:alnum:]_])process\.env\.([[:alnum:]_]+)($|[^[:alnum:]_])/\1env.get('\''\2'\'')\3/g
' {} \;
sed: {}: No such file or directory
I want only the files containing process.env.SOMETHING to be edited.
Work with pipes. xargs comes handy:
find ... -print |
xargs -d '\n' grep -l 'regex' |
xargs -d '\n' sed 'stuff'
xargs: illegal option -- d
You can:
install GNU xargs
install GNU parallel
write a bash loop to read the files line by line, see https://mywiki.wooledge.org/BashFAQ/001
make sure your files do not have spaces or tabs or newlines in filnames and just remove -d '\n' option.
Suggesting to reverse order of commands. sed on filtered list of files.
Files filter is combination of grep filter on find filter: grep -l "process\.env\." $(find . -type f -name '*.js' -not -path './node_modules/*').
sed -i .bak -E '
1i\
const env = require('\''env-var'\'');
s/(^|[^[:alnum:]_])process\.env\.([[:alnum:]_]+) \|\| ([[:alnum:]_]+)($|[^[:alnum:]_])/\1env.get('\''\2'\'').default('\''\3'\'')\4/g
s/(^|[^[:alnum:]_])process\.env\.([[:alnum:]_]+)($|[^[:alnum:]_])/\1env.get('\''\2'\'')\3/g
' $(grep -l "process\.env\." $(find . -type f -name '*.js' -not -path './node_modules/*'))
I would like to find large strings in a directory files and report them:
awk 'length>50' /home/* -exec ls -l {} ';'
Thanks in advance
You need find for that, e.g:
find . -type f -exec grep -Eq '.{50}' {} \; \
-exec ls -l {} +
In GNU find -exec ls -l {} + could be replaced with just -ls.
And if long output is not necessary (requires GNU grep):
grep -Erl '.{50}' .
If your file names don't contain spaces then with POSIX tools:
grep -El '.{50}' /home/* | xargs ls -l
otherwise with GNU tools:
grep -ElZ '.{50}' /home/* | xargs -0 ls -l
I am writing a command using find, grep and sort to display a sorted list of all files that contain 'some-text'.
I was unable to figure out the command.
Here is my attempt:
$find . -type f |grep -l "some-text" | sort
but it didn't work.
You need to use something like XARGS so that the content of each file passed through the pipe | is made available for grep.
XARGS: converts input from standard input into arguments to a command
In my case, I have files1,2,3 and they contain the word test. This will do it.
za:tmp za$ find . -type f | xargs grep -l "test" | sort
./file1.txt
./file2.txt
./file3.txt
or
za:tmp za$ find . -type f | xargs grep -i "test" | sort
./file1.txt:some test string
./file2.txt:some test string
./file3.txt:some test string
You can use it in any unix:
find . -type f -exec sh -c 'grep "some text" {} /dev/null > /dev/null 2>&1' \; -a -print 2> /dev/null|sort
A more optimized solution that works only with GNU-grep:
find . -type f -exec grep -Hq "some-text" {} \; -a -print 2> /dev/null|sort
I'm using command find to recursively browse through directory tree, counting files, sizes, etc...
Now I need to get directory depth of each file.
Is there any portable way for both FreeBSD and CentOS?
I know that find is able to prinf actual directory depth but sadly this works only on CentOS, not FreeBSD.
Additionaly - I need to keep standard find output OR put directory depth on the beginning of output and cut it from there.
You can count the / in path :
$ find . -type f -exec bash -c 'echo '{}' | grep -o / | wc -l' \;
Or with file names :
$ mkdir -p one/two/three four/five && touch file one/two/file one/two/three/file
$ find . -type f -exec bash -c 'echo -n '{}' :; echo '{}' | grep -o / | wc -l' \;
./file :1
./one/two/file :3
./one/two/three/file :4
Try this:
find . -type d -exec bash -c 'echo $(tr -cd / <<< "$1"|wc -c):$1' -- {} \; | sort -n | tail -n 1 | awk -F: '{print $1, $2}'
I'm using command from this topic to view all file extensions in directory and all subdirectories.
find . -type f -name '*.*' | sed 's|.*\.||' | sort -u
How do I can count number of appearance for each extension?
Like:
png: 140
Like this, using uniq with the -c, --count flag:
find . -type f -name '*.*' | sed 's|.*\.||' | sort | uniq -c