Combine grep -v with grep -r? - grep

I want to remove an entire line of text from all files in a given directory. I know I can use grep -v foo filename to do this one file at a time. And I know I can use grep -r foo to search recursively through a directory. How do I combine these commands to remove a given line of text from all files in a directory?

The UNIX command to find files is named find, not grep. Forget you ever heard of grep -r as it's just a bad idea, here's the right way to find files and perform some action on them:
find . -type f -print | xargs sed -i '/badline/d'

Try something like:
grep -vlre 'foo' . | xargs sed -i 's/pattern/replacement/g'
Broken down:
grep:
-v 'Inverse match'
-l 'Show filename'
-r 'Search recursively'
-e 'Extended pattern search'
xargs: For each entry perform
sed -i: replace inline

I think this would work:
grep -ilre 'Foo' . | xargs sed -i 'extension' 'Foo/d'
Where 'extension' refers to the addition to the file name. It will make a copy of the original file with the extension you designated and the modified file will have the original filename. I added -i in case you require it to be case insensitive.
modified file1 becomes "file1"
original file1 becomes "file1extension"
invalid command code ., despite escaping periods, using sed
One of the responses suggests that the newer version of sed's -i option in OSX is slightly different so you need to add an extension. The file is being interpreted as a command, which is why you are seeing that error.

Related

using grep command to get spectfic word [LINUX]

I have a test.txt file with links for example:
google.com?test=
google.com?hello=
and this code
xargs -0 -n1 -a FUZZvul.txt -d '\n' -P 20 -I % curl -ks1L '%/?=DarkLotus' | grep -a 'DarkLotus'
When I type a specific word, such as DarkLotus, in the terminal, it checks the links in the file and it brings me the word which is reflected in the links i provided in the test file
There is no problem here, the problem is that I have many links, and when the result appears in the terminal, I do not know which site reflected the DarkLotus word.
How can i do it?
Try -n option. It shows the line number of file with the matched line.
Best Regards,
Haridas.
I'm not sure what you are up to there, but can you invert it? grep by default prints matching lines. The problem here is you are piping the input from the stdout of the previous commands into grep, and that can lack context at grep. Since you have a file to work with:
$ grep 'DarkLotus' FUZZvul.txt
If your intention is to also follow the link then it might be easier to write a bash script:
#!/bin/bash
for line in `grep 'DarkLotus FUZZvul.txt`
do
link=# extract link from line
echo ${link}
curl -ks1L ${link}
done
Then you could make your script accept user input:
#/bin/bash
word="${0}"
for line in `grep ${word} FUZZvul.txt`
...
and then
$ my_link_getter "DarkLotus"
https://google?somearg=DarkLotus
...
And then you could make the txt file a parameter.
etc.

Using grep to find a string that starts with a character with numbers after

Okay I have a file that contains numbers like this:
L21479
What I am trying to do is use grep (or a similar tool) to find all the strings in a file that have the format:
L#####
The # will be the number. SO an L followed by 5 numbers.
Is this even possible in grep? Should I load the file and perform regex?
You can do this with grep, for example with the following command:
grep -E -o 'L[0-9]{5}' name_of_file
For example, given a file with the text:
kasdhflkashl143112343214L232134614
3L1431413543454L2342L3523269ufoidu
gl9983ugsdu8768IUHI/(JHKJASHD/(888
The command above will output:
L23213
L14314
L35232
If it is just in a single file, you can do something along the lines of:
grep -e 'L[0-9]{5}' filename
If you need to search all files in a directory for these strings:
find . -type f | xargs grep -e 'L[0-9]{5}'

grep with --include and --exclude

I want to search for a string foo within the app directory, but excluding any file which contains migrations in the file name. I expected this grep command to work
grep -Ir --include "*.py" --exclude "*migrations*" foo app/
The above command seems to ignore the --exclude filter. As an alternative, I can do
grep -Ir --include "*.py" foo app/ | grep -v migrations
This works, but this loses highlighting of foo in the results. I can also bring find into the mix and keep my highlighting.
find app/ -name "*.py" -print0 | xargs -0 grep --exclude "*migrations*" foo
I'm just wondering if I'm missing something about the combination of command line parameters to grep or if they simply don't work together.
I was looking for a term on a .py file, but didn't want migration files to be scanned, so what I found (for grep 2.10) was the following (I hope this helps):
grep -nR --include="*.py" --exclude-dir=migrations whatever_you_are_looking_for .
man grep says:
--include=GLOB
Search only files whose base name matches GLOB (using wildcard matching as described under
--exclude).
because it says "only" there, i'm guessing that your --include statment is overriding your --exclude statement.

How can I have grep not print out 'No such file or directory' errors?

I'm grepping through a large pile of code managed by git, and whenever I do a grep, I see piles and piles of messages of the form:
> grep pattern * -R -n
whatever/.git/svn: No such file or directory
Is there any way I can make those lines go away?
You can use the -s or --no-messages flag to suppress errors.
-s, --no-messages suppress error messages
grep pattern * -s -R -n
If you are grepping through a git repository, I'd recommend you use git grep. You don't need to pass in -R or the path.
git grep pattern
That will show all matches from your current directory down.
Errors like that are usually sent to the "standard error" stream, which you can pipe to a file or just make disappear on most commands:
grep pattern * -R -n 2>/dev/null
I have seen that happening several times, with broken links (symlinks that point to files that do not exist), grep tries to search on the target file, which does not exist (hence the correct and accurate error message).
I normally don't bother while doing sysadmin tasks over the console, but from within scripts I do look for text files with "find", and then grep each one:
find /etc -type f -exec grep -nHi -e "widehat" {} \;
Instead of:
grep -nRHi -e "widehat" /etc
I usually don't let grep do the recursion itself. There are usually a few directories you want to skip (.git, .svn...)
You can do clever aliases with stances like that one:
find . \( -name .svn -o -name .git \) -prune -o -type f -exec grep -Hn pattern {} \;
It may seem overkill at first glance, but when you need to filter out some patterns it is quite handy.
Have you tried the -0 option in xargs? Something like this:
ls -r1 | xargs -0 grep 'some text'
Use -I in grep.
Example: grep SEARCH_ME -Irs ~/logs.
I redirect stderr to stdout and then use grep's invert-match (-v) to exclude the warning/error string that I want to hide:
grep -r <pattern> * 2>&1 | grep -v "No such file or directory"
I was getting lots of these errors running "M-x rgrep" from Emacs on Windows with /Git/usr/bin in my PATH. Apparently in that case, M-x rgrep uses "NUL" (the Windows null device) rather than "/dev/null". I fixed the issue by adding this to .emacs:
;; Prevent issues with the Windows null device (NUL)
;; when using cygwin find with rgrep.
(defadvice grep-compute-defaults (around grep-compute-defaults-advice-null-device)
"Use cygwin's /dev/null as the null-device."
(let ((null-device "/dev/null"))
ad-do-it))
(ad-activate 'grep-compute-defaults)
One easy way to make grep return zero status all the time is to use || true
→ echo "Hello" | grep "This won't be found" || true
→ echo $?
0
As you can see the output value here is 0 (Success)

Automatically ignore files in grep

Is there any way I could use grep to ignore some files when searching something, something equivalent to svnignore or gitignore? I usually use something like this when searching source code.
grep -r something * | grep -v ignore_file1 | grep -v ignore_file2
Even if I could set up an alias to grep to ignore these files would be good.
--exclude option on grep will also work:
grep perl * --exclude=try* --exclude=tk*
This searches for perl in files in the current directory excluding files beginning with try or tk.
You might also want to take a look at ack which, among many other features, by default does not search VCS directories like .svn and .git.
find . -path ./ignore -prune -o -exec grep -r something {} \;
What that does is find all files in your current directory excluding the directory (or file) named "ignore", then executes the command grep -r something on each file found in the non-ignored files.
Use shell expansion
shopt -s extglob
for file in !(file1_ignore|file2_ignore)
do
grep ..... "$file"
done
I thinks grep does not have filename filtering.
To accomplish what you are trying to do, you can combine find, xargs, and grep commands.
My memory is not good, so the example might not work:
find -name "foo" | xargs grep "pattern"
Find is flexible, you can use wildcards, ignore case, or use regular expressions.
You may want to read manual pages for full description.
after reading next post, apparently grep does have filename filtering.
Here's a minimalistic version of .gitignore. Requires standard utils: awk, sed (because my awk is so lame), egrep:
cat > ~/bin/grepignore #or anywhere you like in your $PATH
egrep -v "`awk '1' ORS=\| .grepignore | sed -e 's/|$//g' ; echo`"
^D
chmod 755 ~/bin/grepignore
cat >> ./.grepignore #above set to look in cwd
ignorefile_1
...
^D
grep -r something * | grepignore
grepignore builds a simple alternation clause:
egrep -v ignorefile_one|ignorefile_two
not incredibly efficient, but good for manual use

Resources