I use grep -L to get a list of files that do not contain a certain string. How can I see the content of those files? Just like:
grep -L "pattern" | cat
You can use xargs:
grep -L "pattern" | xargs cat
As read in man xargs --> build and execute command lines from standard input. So it will cat to those file names that grep -L returns.
You can use cat and use the output of grep -L...
cat $(grep -L "pattern" *.files )
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.
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
How can I show the count of a grep per file when searching in various files at once via regex
grep -c -? somestring log*
so, if I have n files log_1, log_2, log_3, .., log_n and I grep -c for somestring, then I would like the output to be
log_1 'result of grep -c in file log_1'
log_2 'result of grep -c in file log_2'
Can I do this with a grep command or do I have to construct some sort of loop?
From almas shaik. works great
grep -cH <searchString> <files>
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.