How can I grep for the result of find within another pattern?
That's how I get all filenames with a certain pattern (in my case ending with "ext1")
find . -name *ext1 -printf "%f\n"
And then I want to grep for these filenames with another pattern (in my case ending on "ext2"):
grep -r '[filname]' *ext2
I tried with
find . -name *ext1 -printf "%f\n" | xargs grep -r *ext2
But this only makes grep tell me that it can not find the files found by find.
You would tell grep that the patterns are in a file with the -f option, and use the "stdin filename" -:
find ... | grep -r -f - *ext2
Related
So far, I have this command on my terminal and it doesn't do anything.
Essentially it's to look for any file that contains the word bango and move it to another directory.
grep -r ".*bango.*" /Users/user/Desktop/drums | xargs mv /Users/user/Desktop/bango
Grep has a function to list the filename only you should use that to list the name of the files.
Also xargs can build commands with positional arguments.
Try to use
grep -rlE ".*bango.*" /Users/user/Desktop/drums | xargs -I # mv # /Users/user/Desktop/bango
The option -E allows to use regular expressions.
However, a regular expression is not needed, you can activate a fast grep algorithm for fixed strings:
grep -rlF "bango" /Users/user/Desktop/drums | xargs -I # mv # /Users/user/Desktop/bango
I often have to look for specific strings in a big set of log files with grep. And I get lots of results, on what I must scroll a lot.
Today, the results of grep list the results in alphabetical order. I would like to have my grep results reversed ordered by time, like a ls -ltr would do.
I know I could take the result of ls -ltr and grep file by file. I do it like this:
ls -ltr ${my_log_dir}\* | awk '{print $9}' |xargs grep ${my_pattern}
But I wonder: Is there a simpler way?
PS: I'm using ksh on AIX.
The solution I found (thanks to Fedorqui) was to simply use ls -tr. It assumes the results are passed in the right order through the | to xargs allowing then to do the grep.
My misconception was that since when I us ls, the results arrive not as a single column list but as a multiple column list, it couldn't work as an input for xargs.
Here is the simplest solution to date, since it avoids any awk parsing:
ls -tr ${my_log_dir}\* | xargs grep ${my_pattern}
I checked, and every result of the ls -t are passed to xargs even though they look not as I expected they would enter easily in it:
srv:/papi/tata $ ls -t
addl ieet rrri
ooij lllr sss
srv:/papi/tata $ ls -t |xargs -I{} echo {}
addl
ieet
rrri
ooij
lllr
sss
This will work too use find command:-
find -type f -print0 | xargs -r0 stat -c %y\ %n | sort -r | awk '{print $4}' | sed "s|^\./||"
-print0 in find to preserve files having special characters(whitespaces, tabs)
Print file status (stat with %y (Time of last modification) and %n (%n File name) with output having new-separated (-c)
Reverse sort the output from previous command. (-r for reverse)
awk '{print $4}' printing only the file-name (can be optimized as needed)
Removing the leading ./ from the file-names.
I want to grep -R a directory but exclude symlinks how dow I do it?
Maybe something like grep -R --no-symlinks or something?
Thank you.
Gnu grep v2.11-8 and on if invoked with -r excludes symlinks not specified on the command line and includes them when invoked with -R.
If you already know the name(s) of the symlinks you want to exclude:
grep -r --exclude-dir=LINK1 --exclude-dir=LINK2 PATTERN .
If the name(s) of the symlinks vary, maybe exclude symlinks with a find command first, and then grep the files that this outputs:
find . -type f -a -exec grep -H PATTERN '{}' \;
The '-H' to grep adds the filename to the output (which is the default if grep is searching recursively, but is not here, where grep is being handed individual file names.)
I commonly want to modify grep to exclude source control directories. That is most efficiently done by the initial find command:
find . -name .git -prune -o -type f -a -exec grep -H PATTERN '{}' \;
For now.. here is how I would exclude symbolic links when using grep
If you want just file names matching your search:
for f in $(grep -Rl 'search' *); do if [ ! -h "$f" ]; then echo "$f"; fi; done;
Explaination:
grep -R # recursive
grep -l # file names only
if [ ! -h "file" ] # bash if not a symbolic link
If you want the matched content output, how about a double grep:
srch="whatever"; for f in $(grep -Rl "$srch" *); do if [ ! -h "$f" ]; then
echo -e "\n## $f";
grep -n "$srch" "$f";
fi; done;
Explaination:
echo -e # enable interpretation of backslash escapes
grep -n # adds line numbers to output
.. It's not perfect of course. But it could get the job done!
If you're using an older grep that does not have the -r behavior described in Aryeh Leib Taurog's answer, you can use a combination of find, xargs and grep:
find . -type f | xargs grep "text-to-search-for"
If you are using BSD grep (Mac) the following works similar to '-r' option of Gnu grep.
grep -OR <PATTERN> <PATH> 2> /dev/null
From man page
-O If -R is specified, follow symbolic links only if they were explicitly listed on the command line.
I want to delete all files which have names containing a specific word, e.g. "car".
So far, I came up with this:
find|grep car
How do I pass the output to rm?
find . -name '*car*' -exec rm -f {} \;
or pass the output of your pipeline to xargs:
find | grep car | xargs rm -f
Note that these are very blunt tools, and you are likely to remove files that you did not intend to remove. Also, no effort is made here to deal with files that contain characters such as whitespace (including newlines) or leading dashes. Be warned.
To view what you are going to delete first, since rm -fr is such a dangerous command:
find /path/to/file/ | grep car | xargs ls -lh
Then if the results are what you want, run the real command by removing the ls -lh, replacing it with rm -fr
find /path/to/file/ | grep car | xargs rm -fr
I like to use
rm -rf $(find . | grep car)
It does exactly what you ask, logically running rm -rf on the what grep car returns from the output of find . which is a list of every file and folder recursively.
You can use ls and grep to find your files and rm -rf to delete the files.
rm -rf $(ls | grep car)
But this is not a good idea to use this command if there is a chance of directories or files, you don't want to delete, having names with the character pattern you are specifying with grep.
You really want to use find with -print0 and rm with --:
find [dir] [options] -print0 | grep --null-data [pattern] | xargs -0 rm --
A concrete example (removing all files below the current directory containing car in their filename):
find . -print0 | grep --null-data car | xargs -0 rm --
Why is this necessary:
-print0, --null-data and -0 change the handling of the input/output from parsed as tokens separated by whitespace to parsed as tokens separated by the \0-character. This allows the handling of unusual filenames (see man find for details)
rm -- makes sure to actually remove files starting with a - instead of treating them as parameters to rm. In case there is a file called -rf and do find . -print0 | grep --null-data r | xargs -0 rm, the file -rf will possibly not be removed, but alter the behaviour of rm on the other files.
This finds a file with matching pattern (*.xml) and greps its contents for matching string (exclude="1") and deletes that file if a match is found.
find . -type f -name "*.xml" -exec grep exclude=\"1\" {} \; -exec rm {} \;
Most of the other solutions presented here have problems with handling file names with spaces in them. Here's a solution that handles spaces properly.
grep -lRZ car . | xargs -0 rm
Notes on arguments used:
-l tells grep to print only filenames
-R enables grep recursive search in subfolders
-Z tells grep to separate results by \0 instead of \n
-0 tells xargs to separate input arguments by \0 instead of whitespace
car is the regular expression to search for
. is the folder where to search
Can also use rm -f to force the removal (as usual).
A bit of necromancy, but you can also use find, grep, and xargs
find . -type f | grep -e "pattern1" -e "pattern2" | xargs rm -rf
^ Find will need some attention to make it work for your needs potentially, such as is a file, mindepth, maxdepth and any globbing.
when find | grep car | xargs rm -f get results:
/path/to/car
/path/to/car copy
some files which contain whitespace will not be removed.
So my answer is:
find | grep car | while read -r line ; do
rm -rf "${line}"
done
So the file contains whitespace could be removed.
find start_dir -iname \*car\* -exec rm -v {} \;
I use:
find . | grep "car" | while read i; do echo $i; rm -f "$i"; done
This works even if there are spaces in the filename as well as in recursive manner, searching for directories as well.
Use rm with wildcard *
rm * will delete all files
rm *.ext will delete all files which have ext as extension
rm word* will delete all files which starts with word.
so far I have gotten this far:
prompt$ find path/to/project -type f | grep -v '*.ori|*.pte|*.uh|*.mna' | xargs dos2unix 2> log.txt
However, the files with extensions .ori, .pte, .uh and .mna still show up.
It is better to leave the excluding to find, see Birei's answer.
The problem with your grep pattern is that you have specified it as a shell glob. By default grep expects basic regular expressions (BRE) as its first argument. So if you replace your grep pattern with: .*\.\(ori\|pte\|uh\|mna\)$ it should work. Or if you would rather use extended regular expressions (ERE), you can enable them with -E. Then you can express the same exclusion like this: .*\.(ori|pte|uh|mna)$.
Full command-line:
find . -type f | grep -vE '.*\.(ori|pte|uh|mna)$'
One way:
find path/to/project *.* -type f ! \( -name '*.ori' -o -name '*.pte' -o -name '*.uh' -o -name '*.mna' \)
| xargs dos2unix 2> log.txt