This question already has answers here:
How do I find files that do not contain a given string pattern?
(18 answers)
Closed 6 years ago.
In bash, how to list files that do NOT contain a given string?
Given that
grep --include=*.c -rlw './' -e "pattern"
return any file that matches the pattern I was expecting that
grep --include=*.c -rlwv './' -e "pattern"
would return any file that does not match the pattern but it just returns all the *.c files regardless of wether they match the pattern.
You can try to use -L option:
grep -L -r -i --include \*.c "pattern" ./
You can use -v option also
grep -v -rwl "pattern"
Related
This question already has answers here:
Using the star sign in grep
(12 answers)
Closed 1 year ago.
I have a file test.txt which contains this text data
# cat test.txt
*#=>*#
if I use grep to check if the string is in the file using this way
# grep "*#=>*#" test.txt
#
it returns nothing..
while if I grep a partial string search
# grep "*#=>" test.txt
# *#=>*#
it works correctly ..
Why in the first case do grep return nothing ?
The asterisk is special to grep, it means "the previous token is repeated zero or more times". So >* would match an empty string or > or >> etc.
To match an asterisk, backslash it in the pattern:
grep '\*#=>\*#' test.txt
(The first asterisk follows no token, so the backslash is optional.)
This question already has answers here:
How do I exclude a directory when using `find`?
(46 answers)
Closed 3 years ago.
I need a bash script to iterate on all files in directory besides one with specific names. Maybe it can be done with help of awk/sed during script execution?
Here is my script, that simply merge all file in directory to one:
#!/bin/bash
(find $DIR_NAME -name app.gz\* | sort -rV | xargs -L1 gunzip -c 2> /dev/null || :)
How can I add some $DIR_NAME to list, and don`t iterate over them?
Put the names of the files to be excluded into a file, say "blacklist.txt", one filename per line. Then use
... | grep -F -f blacklist.txt | sort ...
to exclude them from the input to xargs.
This question already has answers here:
Delete files with string found in file - Linux cli
(8 answers)
Closed 5 years ago.
I using this command for search all file contain this word . I want to remove all file contain this word in specific directory . grep command perfectly. suggest me how can I used
rm -rf
with below command
grep -l -r -i "Pending" . | grep -n . | wc -l
This could be done by using the l flag and piping the filenames to xargs:
-l
(The letter ell.) Write only the names of files containing selected
lines to standard output. Pathnames are written once per file searched.
If the standard input is searched, a pathname of (standard input) will
be written, in the POSIX locale. In other locales, standard input may be
replaced by something more appropriate in those locales.
grep -l -r 'Pending' . | xargs rm
The above will delete all files in the current directory containing the word Pending.
I would like to know how to search from the command line for a string in various files of type .rb.
And replace:
.delay([ANY OPTIONAL TEXT FOR DELETION]).
with
.delay.
Besides sed an awk are there any other command line tools included in the OS that are better for the task?
Status
So far I have the following regular expression:
.delay\(*.*\)\.
I would like to know how to match only the expression ending on the first closing parenthesis? And avoid replacing:
.delay([ANY OPTIONAL TEXT FOR DELETION]).sometext(param)
Thanks in advance!
If you need to find and replace text in files - sed seems to be the best command line solution.
Search for a string in the text file and replace:
sed -i 's/PATTERN/REPLACEMENT/' file.name
Or, if you need to process multiple occurencies of PATTERN in file, add g key
sed -i 's/PATTERN/REPLACEMENT/g' file.name
For multiple files processing - redirect list of files to sed:
echo "${filesList}" | xargs sed -i ...
You can use find to generate your list of files, and xargs to run sed over the result:
find . -type f -print | xargs sed -i 's/\.delay.*/.delay./'
find will generate a list of files contained in your current directory (., although you can of course pass a different directory), xargs will read that list and then run sed with the list of files as an argument.
Instead of find, which here generates a list of all files, you could use something like grep to generate a list of files that contain a specific term. E.g.:
grep -rl '\.delay' | xargs sed -i ...
For the part of the question where you want to only match and replace until the first ) and not include a second pair of (), here is how to change your regex:
.delay\(*.*\)\.
->
\.delay\([^\)]*\)
I.e. match "actual dot, delay, brace open, everything but brace close and brace close".
E.g. using sed:
>echo .delay([ANY OPTIONAL TEXT FOR DELETION]).sometext(param) | sed -E "s/\.delay\([^\)]*\)/.delay/"
.delay.sometext(param)
I recommend to use grep for finding the right files:
grep -rl --include "*.rb" '\.delay' .
Then feed the list into xargs, as recommended by other answers.
Credits to the other answers for providing a solution for feeding multiple files into sed.
This question already has answers here:
How to find all files containing specific text (string) on Linux?
(54 answers)
Closed 8 years ago.
I need to grep for a particular port number from a huge set of files.
I am using a command:
find . |xargs grep "9461"
But it does not finds all the occurrences for number 9461.
Can anyone suggest a better unix/linux command to do so.
The kind of files it gets is : x.log, y.txt,z.htm, a.out etc files
But it was not able to get abc.conf files
You surely have some reason for using find in combination with grep, but just in case:
You can replace your command by:
grep -r "9461" .
and if you want even line numbers
grep -rn "9461" .
As JonathanLefflero commented, there is also the option -e that make grep match againt a regular expression, so, the ultimate command would be
grep -rne 9461
You should take a look on grep man page
A final note, you should check if what you want to grep is "9461" or 9461 without "".