I am using the code below to grep some string:
grep 'string' *.log | grep -v 'string1'
I am getting output in particular file. My requirement is to extract that file name to a variable. How I can do that effectively?
In general, you can capture the output of any command into a shell variable via command substitution like this:
variable=$(command arg1 arg2)
This is appropriate for your particular case if you are sure that there will be only one file name produced by the grep pipeline. In that case, you capture its name into shell variable fname via:
fname=$(grep -lZ string *.log | xargs -0 grep -lv string1)
This is safe for difficult file names because, via the -Z and -0 options, we use NUL-separated lists. The -l option to grep is useful here because it suppresses the normal grep output and just prints the file names.
If there might be multiple file matches, then, if you can use an advanced shell like bash, try:
grep -lZ string *.log | xargs -0 grep -lvZ string1 | while IFS= read -r -d $'\0' fname
do
# Process file "$fname"
done
This is also safe for difficult file names because, throughout the pipeline, it uses NUL-separated lists.
For a POSIX shell, read works with newline-separated input. To make the above safe for difficult file names, the -d option is used which is supported by bash, zsh, and other advanced shells.
use command
basename "filePath" "fileExtension"
ex: basename /home/john/xyz.txt .txt
output: xyz
Related
I am facing a problem with a script I need to use for log analysis; let me explain the question:
I have a gzipped file like:
5555_prova.log.gz
Inside the file there are mali lines of log like this one:
2018-06-12 03:34:31 95.245.15.135 GET /hls.playready.vod.mediasetpremium/farmunica/2018/06/218742_163f10da04c7d2/hlsrc/w12/21.ts
I need a script read the gzipped log file which is capable to output on the stdout a modified log line like this one:
5555 2018-06-12 03:34:31 95.245.15.135 GET /hls.playready.vod.mediasetpremium/farmunica/2018/06/218742_163f10da04c7d2/hlsrc/w12/21.ts
As you can see the line of log now start with the number read from the gzip file name.
I need this new line to feed a logstash data crunching chain.
I have tried with a script like this:
echo "./5555_prova.log.gz" | xargs -ISTR -t -r sh -c "gunzip -c STR | awk '{$0="5555 "$0}' "
this is not exactly what I need (the prefix is static and not captured with a regular expression from the file name) but even with this simplified version I receive an error:
sh -c gunzip -c ./5555_prova.log.gz | awk '{-bash=5555 -bash}'
-bash}' : -c: line 0: unexpected EOF while looking for matching `''
-bash}' : -c: line 1: syntax error: unexpected end of file
As you can see from the above output the $0 is no more the whole line passed via pipe to awk but is a strange -bash.
I need to use xargs because the list of gzipped file is fed the the command line from an another tool (i.e. an instantiated inotifywait listening to a directory where the files are written via ftp).
What I am missing? do you have some suggestions to point me in the right direction?
Regards,
S.
Trying to following the #Charles Duffy suggestion I have written this code:
#/bin/bash
#
# Usage: sendToLogstash.sh [pattern]
#
# Executes a command whenever files matching the pattern are closed in write
# mode or moved to. "{}" in the command is replaced with the matching filename (via xargs).
# Requires inotifywait from inotify-tools.
#
# For example,
#
# whenever.sh '/usr/local/myfiles/'
#
#
DIR="$1"
PATTERN="\.gz$"
script=$(cat <<'EOF'
awk -v filename="$file" 'BEGIN{split(filename,array,"_")}{$0=array[1] OFS $0} 1' < $(gunzip -dc "$DIR/$file")
EOF
)
inotifywait -q --format '%f' -m -r -e close_write -e moved_to "$DIR" \
| grep --line-buffered $PATTERN | xargs -I{} -r sh -c "file={}; $script"
But I got the error:
[root#ms-felogstash ~]# ./test.sh ./poppo
gzip: /1111_test.log.gz: No such file or directory
gzip: /1111_test.log.gz: No such file or directory
sh: $(gunzip -dc "$DIR/$file"): ambiguous redirect
Thanks for your help, I feel very lost writing bash scripts.
Regards,
S.
EDIT: Also in case you are dealing with multiple .gz files and want to print their content along with their file names(first column _ delimited) then following may help you.
for file in *.gz; do
awk -v filename="$file" 'BEGIN{split(filename,array,"_")}{$0=array[1] OFS $0} 1' <(gzip -dc "$file")
done
I haven't tested your code(couldn't completely understand also), so trying to give here a way like in case your code could pass file name to awk then it will be pretty simple to append the file's first digits like as follows(just an example).
awk 'FNR==1{split(FILENAME,array,"_")} {$0=array[1] OFS $0} 1' 5555_prova.log_file
So here I am taking FILENAME out of the box variable for awk(only in first line of file) and then by splitting it into array named array and then adding it in each line of the file.
Also wrap "gunzip -c STR this with ending " which seems to be missing before you pass its output to awk too.
NEVER, EVER use xargs -I with a string substituted into sh -c (or bash -c or any other context where that string is interpreted as code). This allows malicious filenames to run arbitrary commands -- think about what happens if someone runs touch $'$(rm -rf ~)\'$(rm -rf ~)\'.gz', and gets that file into your log.
Instead, let xargs append arguments after your script text, and write your script to iterate over / read those arguments as data, rather than having them substituted into code.
To show how to use xargs safely (well, safely if we assume that you've filtered out filenames with literal newlines):
# This way you don't need to escape the quotes in your script by hand
script=$(cat <<'EOF'
for arg; do gunzip -c <"$arg" | awk '{$0="5555 "$0}'; done
EOF
)
# if you **did** want to escape them by hand, it would look like this:
# script='for arg; do gunzip -c <"$arg" | awk '"'"'{$0="5555 "$0}'"'"'; done'
echo "./5555_prova.log.gz" | xargs -d $'\n' sh -c "$script" _
To be safer with all possible filenames, you'd instead use:
printf '%s\0' "./5555_prova.log.gz" | xargs -0 sh -c "$script" _
Note the use of NUL-delimited input (created with printf '%s\0') and xargs -0 to consume it.
How can I perform an operation for each item listed by grep individually?
Background:
I use grep to list all files containing a certain pattern:
grep -l '<pattern>' directory/*.extension1
I want to delete all listed files but also all files having the same file name but a different extension: .extension2.
I tried using the pipe, but it seems to take the output of grep as a whole.
In find there is the -exec option, but grep has nothing like that.
If I understand your specification, you want:
grep --null -l '<pattern>' directory/*.extension1 | \
xargs -n 1 -0 -I{} bash -c 'rm "$1" "${1%.*}.extension2"' -- {}
This is essentially the same as what #triplee's comment describes, except that it's newline-safe.
What's going on here?
grep with --null will return output delimited with nulls instead of newline. Since file names can have newlines in them delimiting with newline makes it impossible to parse the output of grep safely, but null is not a valid character in a file name and thus makes a nice delimiter.
xargs will take a stream of newline-delimited items and execute a given command, passing as many of those items (one as each parameter) to a given command (or to echo if no command is given). Thus if you said:
printf 'one\ntwo three \nfour\n' | xargs echo
xargs would execute echo one 'two three' four. This is not safe for file names because, again, file names might contain embedded newlines.
The -0 switch to xargs changes it from looking for a newline delimiter to a null delimiter. This makes it match the output we got from grep --null and makes it safe for processing a list of file names.
Normally xargs simply appends the input to the end of a command. The -I switch to xargs changes this to substitution the specified replacement string with the input. To get the idea try this experiment:
printf 'one\ntwo three \nfour\n' | xargs -I{} echo foo {} bar
And note the difference from the earlier printf | xargs command.
In the case of my solution the command I execute is bash, to which I pass -c. The -c switch causes bash to execute the commands in the following argument (and then terminate) instead of starting an interactive shell. The next block 'rm "$1" "${1%.*}.extension2"' is the first argument to -c and is the script which will be executed by bash. Any arguments following the script argument to -c are assigned as the arguments to the script. This, if I were to say:
bash -c 'echo $0' "Hello, world"
Then Hello, world would be assigned to $0 (the first argument to the script) and inside the script I could echo it back.
Since $0 is normally reserved for the script name I pass a dummy value (in this case --) as the first argument and, then, in place of the second argument I write {}, which is the replacement string I specified for xargs. This will be replaced by xargs with each file name parsed from grep's output before bash is executed.
The mini shell script might look complicated but it's rather trivial. First, the entire script is single-quoted to prevent the calling shell from interpreting it. Inside the script I invoke rm and pass it two file names to remove: the $1 argument, which was the file name passed when the replacement string was substituted above, and ${1%.*}.extension2. This latter is a parameter substitution on the $1 variable. The important part is %.* which says
% "Match from the end of the variable and remove the shortest string matching the pattern.
.* The pattern is a single period followed by anything.
This effectively strips the extension, if any, from the file name. You can observe the effect yourself:
foo='my file.txt'
bar='this.is.a.file.txt'
baz='no extension'
printf '%s\n'"${foo%.*}" "${bar%.*}" "${baz%.*}"
Since the extension has been stripped I concatenate the desired alternate extension .extension2 to the stripped file name to obtain the alternate file name.
If this does what you want, pipe the output through /bin/sh.
grep -l 'RE' folder/*.ext1 | sed 's/\(.*\).ext1/rm "&" "\1.ext2"/'
Or if sed makes you itchy:
grep -l 'RE' folder/*.ext1 | while read file; do
echo rm "$file" "${file%.ext1}.ext2"
done
Remove echo if the output looks like the commands you want to run.
But you can do this with find as well:
find /path/to/start -name \*.ext1 -exec grep -q 'RE' {} \; -print | ...
where ... is either the sed script or the three lines from while to done.
The idea here is that find will ... well, "find" things based on the qualifiers you give it -- namely, that things match the file glob "*.ext", AND that the result of the "exec" is successful. The -q tells grep to look for RE in {} (the file supplied by find), and exit with a TRUE or FALSE without generating any of its own output.
The only real difference between doing this in find vs doing it with grep is that you get to use find's awesome collection of conditions to narrow down your search further if required. man find for details. By default, find will recurse into subdirectories.
You can pipe the list to xargs:
grep -l '<pattern>' directory/*.extension1 | xargs rm
As for the second set of files with a different extension, I'd do this (as usual use xargs echo rm when testing to make a dry run; I haven't tested it, it may not work correctly with filenames with spaces in them):
filelist=$(grep -l '<pattern>' directory/*.extension1)
echo $filelist | xargs rm
echo ${filelist//.extension1/.extension2} | xargs rm
Pipe the result to xargs, it will allow you to run a command for each match.
I am having trouble with this simple task:
cat file | grep -E ^[0-9]+$ > file_grep
diff file file_grep
Problem is, I want to do this without file_grep
I have tried:
diff file `cat file | grep -E ^[0-9]+$`
and
diff file "`cat file | grep -E ^[0-9]+$`"
and a few other combinations :-) but I can't get it to work.
I always get an error, when the diff gets extra argument which is content of file filtered by grep.
Something similar always worked for me, when I wanted to echo command outputs from within a script like this (using backtick escapes):
echo `ls`
Thanks
If you're using bash:
diff file <(grep -E '^[0-9]+$' file)
The <(COMMAND) sequence expands to the name of a pseudo-file (such as /dev/fd/63) from which you can read the output of the command.
But for this particular case, ruakh's solution is simpler. It takes advantage of the fact that - as an argument to diff causes it to read its standard input. The <(COMMAND) syntax becomes more useful when both arguments to diff are command output, such as:
diff <(this_command) <(that_command)
The simplest approach is:
grep -E '^[0-9]+$' file | diff file -
The hyphen - as the filename is a specific notation that tells diff "use standard input"; it's documented in the diff man-page. (Most of the common utilities support the same notation.)
The reason that backticks don't work is that they capture the output of a command and pass it as an argument. For example, this:
cat `echo file`
is equivalent to this:
cat file
and this:
diff file "`cat file | grep -E ^[0-9]+$`"
is equivalent to something like this:
diff file "123
234
456"
That is, it actually tries to pass 123234345 (plus newlines) as a filename, rather than as the contents of a file. Technically, you could achieve the latter by using Bash's "process substitution" feature that actually creates a sort of temporary file:
diff file <(cat file | grep -E '^[0-9]+$')
but in your case it's not needed, because of diff's support for -.
grep -E '^[0-9]+$' file | diff - file
where - means "read from standard input".
Try process substitution:
$ diff file <(grep -E "^[0-9]+$" file)
From the bash manpage:
Process Substitution
Process substitution is supported on systems that support named pipes (FIFOs) or the /dev/fd method of
naming open files. It takes the form of <(list) or >(list). The process list is run with its input or
output connected to a FIFO or some file in /dev/fd. The name of this file is passed as an argument to
the current command as the result of the expansion. If the >(list) form is used, writing to the file
will provide input for list. If the <(list) form is used, the file passed as an argument should be read
to obtain the output of list.
In bash, the syntax is
diff file <(cat file | grep -E ^[0-9]+$)
The following is a simple Bash command line:
grep -li 'regex' "filename with spaces" "filename"
No problems. Also the following works just fine:
grep -li 'regex' $(<listOfFiles.txt)
where listOfFiles.txt contains a list of filenames to be grepped, one
filename per line.
The problem occurs when listOfFiles.txt contains filenames with
embedded spaces. In all cases I've tried (see below), Bash splits the
filenames at the spaces so, for example, a line in listOfFiles.txt
containing a name like ./this is a file.xml ends up trying to run
grep on each piece (./this, is, a and file.xml).
I thought I was a relatively advanced Bash user, but I cannot find a
simple magic incantation to get this to work. Here are the things I've
tried.
grep -li 'regex' `cat listOfFiles.txt`
Fails as described above (I didn't really expect this to work), so I
thought I'd put quotes around each filename:
grep -li 'regex' `sed -e 's/.*/"&"/' listOfFiles.txt`
Bash interprets the quotes as part of the filename and gives "No such
file or directory" for each file (and still splits the filenames with
blanks)
for i in $(<listOfFiles.txt); do grep -li 'regex' "$i"; done
This fails as for the original attempt (that is, it behaves as if the
quotes are ignored) and is very slow since it has to launch one 'grep'
process per file instead of processing all files in one invocation.
The following works, but requires some careful double-escaping if
the regular expression contains shell metacharacters:
eval grep -li 'regex' `sed -e 's/.*/"&"/' listOfFiles.txt`
Is this the only way to construct the command line so it will
correctly handle filenames with spaces?
Try this:
(IFS=$'\n'; grep -li 'regex' $(<listOfFiles.txt))
IFS is the Internal Field Separator. Setting it to $'\n' tells Bash to use the newline character to delimit filenames. Its default value is $' \t\n' and can be printed using cat -etv <<<"$IFS".
Enclosing the script in parenthesis starts a subshell so that only commands within the parenthesis are affected by the custom IFS value.
cat listOfFiles.txt |tr '\n' '\0' |xargs -0 grep -li 'regex'
The -0 option on xargs tells xargs to use a null character rather than white space as a filename terminator. The tr command converts the incoming newlines to a null character.
This meets the OP's requirement that grep not be invoked multiple times. It has been my experience that for a large number of files avoiding the multiple invocations of grep improves performance considerably.
This scheme also avoids a bug in the OP's original method because his scheme will break where listOfFiles.txt contains a number of files that would exceed the buffer size for the commands. xargs knows about the maximum command size and will invoke grep multiple times to avoid that problem.
A related problem with using xargs and grep is that grep will prefix the output with the filename when invoked with multiple files. Because xargs invokes grep with multiple files one will receive output with the filename prefixed, but not for the case of one file in listOfFiles.txt or the case of multiple invocations where the last invocation contains one filename. To achieve consistent output add /dev/null to the grep command:
cat listOfFiles.txt |tr '\n' '\0' |xargs -0 grep -i 'regex' /dev/null
Note that was not an issue for the OP because he was using the -l option on grep; however it is likely to be an issue for others.
This works:
while read file; do grep -li dtw "$file"; done < listOfFiles.txt
With Bash 4, you can also use the builtin mapfile function to set an array containing each line and iterate on this array:
$ tree
.
├── a
│ ├── a 1
│ └── a 2
├── b
│ ├── b 1
│ └── b 2
└── c
├── c 1
└── c 2
3 directories, 6 files
$ mapfile -t files < <(find -type f)
$ for file in "${files[#]}"; do
> echo "file: $file"
> done
file: ./a/a 2
file: ./a/a 1
file: ./b/b 2
file: ./b/b 1
file: ./c/c 2
file: ./c/c 1
Though it may overmatch, this is my favorite solution:
grep -i 'regex' $(cat listOfFiles.txt | sed -e "s/ /?/g")
Do note that if you somehow ended up with a list in a file which has Windows line endings, \r\n, NONE of the notes above about the input file separator $IFS (and quoting the argument) will work; so make sure that the line endings are correctly \n (I use scite to show the line endings, and easily change them from one to the other).
Also cat piped into while file read ... seems to work (apparently without need to set separators):
cat <(echo -e "AA AA\nBB BB") | while read file; do echo $file; done
... although for me it was more relevant for a "grep" through a directory with spaces in filenames:
grep -rlI 'search' "My Dir"/ | while read file; do echo $file; grep 'search\|else' "$ix"; done
I have a file which contains filenames (and the full path to them) and I want to search for a word within all of them.
some pseudo-code to explain:
grep keyword <all files specified in files.txt>
or
cat files.txt > grep keyword
cat files txt | grep keyword
the problem is that I can only get grep to search the filenames, not the contents of the actual files.
cat files.txt | xargs grep keyword
or
grep keyword `cat files.txt`
or (equivalent to previous but harder to mis-read)
grep keyword $(cat files.txt)
should do the trick.
Pitfalls:
If files.txt contains file names with spaces, either solution will malfunction, because "This is a filename.txt" will be interpreted as four files, "This", "is", "a", and "filename.txt". A good reason why you shouldn't have spaces in your filenames, ever.
There are ways around this, but none of them is trivial. (find ... -print0 / xargs -0 is one of them.)
The second (cat) version can result in a very long command line (which might fail when exceeding the limits of your environment). The first (xargs) version handles long input automatically; xargs offers several options to control the details.
Both of the answers from DevSolar work (tested on Linux Ubuntu), but the xargs version is preferable if there may be many files, since it will avoid running into command line length limits.
so:
cat files.txt | xargs grep keyword
is the way to go
tr '\n' '\0' <files.txt | LANG=C xargs -r0 grep -F keyword
tr will delimit names with NUL character so that spaces not significant (note the corresponding -0 option to xargs).
xargs -r will start a single grep process for a "large" number of files, but not start any grep process if there are no files.
LANG=C means use quick routines for matching, rather than slow locale ones
grep -F means use quick string matching rather than slow regular expression matching
bash, ksh & zsh version:
grep keyword $(<files.txt)
Long time when last created a bash shell script, but you could store the result of the first grep (the one finding all filenames) in an array and iterate over it, issuing even more grep commands.
A good starting point should be the bash scripting guide.