I'm trying to grep -R an exact version of apache.
I leave the include and the manuals, because its not necessary to change.
This is the command I'm using:
grep -R 2.2.24 * | grep -v manual | grep -v include | cut -d':' -f 1
I would like to know how to find one result per file?
Thanks
The cut in your command suggests that you're looking to list files matching the specified string.
find . ! -path "*/manual/*" ! -path "*/include/*" -type f -exec grep -Fl 2.2.24 {} \;
should work for you.
Related
I would like to find files in my home directory that start with '~', sort them numerically, print the first five and delete them using find command and pipes in Linux. I have a bash script:
#!/bin/bash
find ~/ -name "~*" | sort -n | head -5 | tee | xargs rm
This works fine for deleting files, but I was expecting tee command to print deleted files to standard output. All this command does is delete files, but there in so output in terminal. What should I add/ change?
Thank you.
You could just use the verbose flag on rm and it will tell you what it's deleting
find ~/ -name "~*" | sort -n | head -5 | xargs rm -v
Use man rm to see the docs
-v, --verbose
explain what is being done
You can use rm -v to print each deleting filename:
find ~ -name '~*' -print0 | sort -zn | head -z -n 5 | xargs -0 rm -v
Also note use -print0 and all corresponding options in sort. head, xargs to address filenames with whitespace and glob characters.
I want to print all the lines that contains "testing.$" from all folder's files.
I am trying the following command, but I am getting a error. What am I missing?
find ./* -type f -exec grep -l testing.$ {} \ | grep -n ^ | grep
^Linha_do_Arquivo: | cut -d: -f2;
find: missing argument: '-exec'
You don't need find for this.
With grep only:
grep -rh "testing.$" .
-h hides the filename in the output. Omit it if you want to output it as well.
I'm trying to get my terminal to return the latest .txt file, with path intact. I've been researching ls, grep, find, and tail, using the '|' functionality of passing results from one utility to the next. The end result would be to have a working path + result that I could pass my text editor.
I've been getting close with tests like this:
find . | grep '.txt$' | tail -1
..but I haven't had luck with grep returning the newest file - is there a flag I'm missing?
Trying to use find & ls isn't exactly working either:
find . -name "*.txt" | ls -lrth
..the ls returns the current directories instead of the results of my find query.
Please help!
You're so very close.
vi "$(find . -name '*.txt' -exec ls -t {} + | head -1)"
find /usr/share -name '*.txt' -printf '%C+ %p\n' | sort -r | head -1 | sed 's/^[^ ]* //'
If you have bash4+
ls -t ./**/*.txt | head -1
edit the latest txt file
vim $(ls -t ./**/*.txt |head -1)
ps: need enabled shopt -s globstar in your .bashrc or .profile...
You can use the stat function to print each file with just the latest modification time and name.
find . -name "*.txt" -exec stat -c "%m %N" {} \; | sort
How do I recursively view a list of files that has one string and specifically doesn't have another string? Also, I mean to evaluate the text of the files, not the filenames.
Conclusion:
As per comments, I ended up using:
find . -name "*.html" -exec grep -lR 'base\-maps' {} \; | xargs grep -L 'base\-maps\-bot'
This returned files with "base-maps" and not "base-maps-bot". Thank you!!
Try this:
grep -rl <string-to-match> | xargs grep -L <string-not-to-match>
Explanation: grep -lr makes grep recursively (r) output a list (l) of all files that contain <string-to-match>. xargs loops over these files, calling grep -L on each one of them. grep -L will only output the filename when the file does not contain <string-not-to-match>.
The use of xargs in the answers above is not necessary; you can achieve the same thing like this:
find . -type f -exec grep -q <string-to-match> {} \; -not -exec grep -q <string-not-to-match> {} \; -print
grep -q means run quietly but return an exit code indicating whether a match was found; find can then use that exit code to determine whether to keep executing the rest of its options. If -exec grep -q <string-to-match> {} \; returns 0, then it will go on to execute -not -exec grep -q <string-not-to-match>{} \;. If that also returns 0, it will go on to execute -print, which prints the name of the file.
As another answer has noted, using find in this way has major advantages over grep -Rl where you only want to search files of a certain type. If, on the other hand, you really want to search all files, grep -Rl is probably quicker, as it uses one grep process to perform the first filter for all files, instead of a separate grep process for each file.
These answers seem off as the match BOTH strings. The following command should work better:
grep -l <string-to-match> * | xargs grep -c <string-not-to-match> | grep '\:0'
Here is a more generic construction:
find . -name <nameFilter> -print0 | xargs -0 grep -Z -l <patternYes> | xargs -0 grep -L <patternNo>
This command outputs files whose name matches <nameFilter> (adjust find predicates as you need) which contain <patternYes>, but do not contain <patternNo>.
The enhancements are:
It works with filenames containing whitespace.
It lets you filter files by name.
If you don't need to filter by name (one often wants to consider all the files in current directory), you can strip find and add -R to the first grep:
grep -R -Z -l <patternYes> | xargs -0 grep -L <patternNo>
find . -maxdepth 1 -name "*.py" -exec grep -L "string-not-to-match" {} \;
This Command will get all ".py" files that don't contain "string-not-to-match" at same directory.
To match string A and exclude strings B & C being present in the same line I use, and quotes to allow search string to contain a space
grep -r <string A> | grep -v -e <string B> -e "<string C>" | awk -F ':' '{print $1}'
Explanation: grep -r recursively filters all lines matching in output format
filename: line
To exclude (grep -v) from those lines the ones that also contain either -e string B or -e string C. awk is used to print only the first field (the filename) using the colon as fieldseparator -F
I am trying to delete erroneous emails based on finding the email address in the file via Linux CLI.
I can get the files with
find . | xargs grep -l email#example.com
But I cannot figure out how to delete them from there as the following code doesn't work.
rm -f | xargs find . | xargs grep -l email#example.com
Solution for your command:
grep -l email#example.com * | xargs rm
Or
for file in $(grep -l email#example.com *); do
rm -i $file;
# ^ prompt for delete
done
For safety I normally pipe the output from find to something like awk and create a batch file with each line being "rm filename"
That way you can check it before actually running it and manually fix any odd edge cases that are difficult to do with a regex
find . | xargs grep -l email#example.com | awk '{print "rm "$1}' > doit.sh
vi doit.sh // check for murphy and his law
source doit.sh
You can use find's -exec and -delete, it will only delete the file if the grep command succeeds. Using grep -q so it wouldn't print anything, you can replace the -q with -l to see which files had the string in them.
find . -exec grep -q 'email#example.com' '{}' \; -delete
I liked Martin Beckett's solution but found that file names with spaces could trip it up (like who uses spaces in file names, pfft :D). Also I wanted to review what was matched so I move the matched files to a local folder instead of just deleting them with the 'rm' command:
# Make a folder in the current directory to put the matched files
$ mkdir -p './matched-files'
# Create a script to move files that match the grep
# NOTE: Remove "-name '*.txt'" to allow all file extensions to be searched.
# NOTE: Edit the grep argument 'something' to what you want to search for.
$ find . -name '*.txt' -print0 | xargs -0 grep -al 'something' | awk -F '\n' '{ print "mv \""$0"\" ./matched-files" }' > doit.sh
Or because its possible (in Linux, idk about other OS's) to have newlines in a file name you can use this longer, untested if works better (who puts newlines in filenames? pfft :D), version:
$ find . -name '*.txt' -print0 | xargs -0 grep -alZ 'something' | awk -F '\0' '{ for (x=1; x<NF; x++) print "mv \""$x"\" ./matched-files" }' > doit.sh
# Evaluate the file following the 'source' command as a list of commands executed in the current context:
$ source doit.sh
NOTE: I had issues where grep could not match inside files that had utf-16 encoding.
See here for a workaround. In case that website disappears what you do is use grep's -a flag which makes grep treat files as text and use a regex pattern that matches any first-byte in each extended character. For example to match Entité do this:
grep -a 'Entit.e'
and if that doesn't work then try this:
grep -a 'E.n.t.i.t.e'
Despite Martin's safe answer, if you've got certainty of what you want to delete, such as in writing a script, I've used this with greater success than any other one-liner suggested before around here:
$ find . | grep -l email#example.com | xargs -I {} rm -rf {}
But I rather find by name:
$ find . -iname *something* | xargs -I {} echo {}
rm -f `find . | xargs grep -li email#example.com`
does the job better. Use `...` to run the command to offer the file names containing email.#example.com (grep -l lists them, -i ignores case) to remove them with rm (-f forcibly / -i interactively).
find . | xargs grep -l email#example.com
how to remove:
rm -f 'find . | xargs grep -l email#example.com'
Quick and efficent. Replace find_files_having_this_text with the text you want to search.
grep -Ril 'find_files_having_this_text' . | xargs rm