UNIX grep command (grep -v grep) - unix

I was going through something and found this which I could not understand,
grep -v grep
What does this signify? I know that -v switch will select all the lines that do not match. But why the second grep?
This is the full command:
ps -ef | grep rsync -avz \
| grep oradata${DAY}_[0-1][0-9] \
| grep -v grep \
| awk '{print $2}' | wc -l

grep when used with ps -ef also outputs the grep used for filtering the output of ps -ef.
grep -v grep means that do not include the grep used for filtering in the command output.
You can also avoid grep in the results by using a regex pattern.
For example, in the following example you won't need a grep -v grep to avoid grep in the output:
ps -ef | grep [r]sync
Here's another example, showing different commands and their output, notice the first one where grep is also in the output whereas in the last two grep is not printed in the output:
$ ps -ef | grep ipython
501 18055 18031 0 12:44AM ttys000 0:00.00 /bin/bash /Users/amit/anaconda/bin/python.app /Users/amit/anaconda/bin/ipython notebook --profile=ocean
501 18056 18055 0 12:44AM ttys000 0:00.85 /Users/amit/anaconda/python.app/Contents/MacOS/python /Users/amit/anaconda/bin/ipython notebook --profile=ocean
501 18067 18031 0 12:44AM ttys000 0:00.00 grep ipython
$ ps -ef | grep ipython | grep -v grep
501 18055 18031 0 12:44AM ttys000 0:00.00 /bin/bash /Users/amit/anaconda/bin/python.app /Users/amit/anaconda/bin/ipython notebook --profile=ocean
501 18056 18055 0 12:44AM ttys000 0:00.85 /Users/amit/anaconda/python.app/Contents/MacOS/python /Users/amit/anaconda/bin/ipython notebook --profile=ocean
$ ps -ef | grep [i]python
501 18055 18031 0 12:44AM ttys000 0:00.00 /bin/bash /Users/amit/anaconda/bin/python.app /Users/amit/anaconda/bin/ipython notebook --profile=ocean
501 18056 18055 0 12:44AM ttys000 0:00.85 /Users/amit/anaconda/python.app/Contents/MacOS/python /Users/amit/anaconda/bin/ipython notebook --profile=ocean

Related

How can pipes and grep and wc be combined to just give a count of the phrase “syntax ok”

How can pipes and grep and wc be combined to just give a count of the phrase “syntax ok”
Something like the following…
cd /usr/IBMIHS/bin/ |
apachectl -t -f /usr/IBMIHS/conf/AAA/httpd.conf |
apachectl -t -f /usr/IBMIHS/conf/AAA/siteAA.conf |
grep "^Syntax OK" | wc
Simply via grouping commands with curly brackets, and use grep -c:
{
apachectl -t -f /usr/IBMIHS/conf/AAA/httpd.conf
apachectl -t -f /usr/IBMIHS/conf/AAA/siteAA.conf
} |& grep -c "Syntax OK"
From man grep
-c, --count
Suppress normal output; instead print a count of matching lines for each input file. With the -v, --invert-match option (see below), count non-matching lines.

How can I ssh into a second server, run a command, and assign the output to a variable?

This is what I am attempting to do:
fromServer=$(ssh -A first.com ssh second.com rpm -qa | grep exampleString)
echo $fromServer
echo does not print anything. If I manually shh into first and then ssh into second then run the command I get output:
ssh first.com
ssh second.com
rpm -qa | grep exampleString
How can I combine these three steps into one line and store the output into a variable?
Use proper quoting or escaping:
fromServer=$(ssh -A first.com 'ssh second.com rpm -qa | grep exampleString')
echo $fromServer
or
fromServer=$(ssh -A first.com ssh second.com rpm -qa \| grep exampleString)
echo $fromServer
% VAR=$(ssh -C user#server ls -la \| grep vim)
% echo $VAR
-rw------- 1 user user 15153 Mar 22 13:45 .vimrc
edit: oooooooh, sneaky, I did not see you were doing two SSH ☺
So then you'll need a bit more quoting, because you don't want to have your pipe being interpreted by first.com. Here's three ways to work around that:
fromServer=$(ssh -A first.com ssh second.com rpm -qa \\\| grep exampleString)
fromServer=$(ssh -A first.com 'ssh second.com rpm -qa \| grep exampleString')
fromServer=$("ssh -A first.com 'ssh second.com rpm -qa | grep exampleString'")
What's happening is that you want to execute:
user#second % rpm -qa | grep exampleString
on the second.com server, so you have to escape the pipe so it's not interpreted by the first.com server:
user#first % ssh second.com rpm -qa \| grep exampleString
or
user#first % ssh second.com 'rpm -qa | grep exampleString'
but then again, you need to have that executed on first.com, from your local workstation, as you still don't want to see the pipe interpreted, you need to add a second layer of escaping/quoting:
user#workstation % ssh first.com "ssh second.com 'rpm -qa | grep exampleString'"
or
user#workstation % ssh first.com 'ssh second.com rpm -qa \| grep exampleString'
and then, once you're sure you get an output you can put that whole command's output in a variable:
VAR=$(ssh first.com "ssh second.com 'rpm -qa | grep exampleString'")
HTH

problem in a shell command

i am trying the following command on the command line
ps -u `id | cut -f2 -d"=" | cut -f1 -d"("` -f | grep ppLSN | awk '{print $9}' | awk '{FS="=";print $2}' | grep KLMN | wc -l
the value of teh command is returned as 7.
but when i am putting the same command inside a script abc_sh like below
ps -u `id | cut -f2 -d"=" | cut -f1 -d"("` -f | grep ppLSN | awk '{print $9}' | awk '{FS="=";print $2}' | grep $XYZ | wc -l
and i am calling the script on the command line as abc_sh XYZ=KLMN and it does not work and returns 0
the problem is with the grep in the command grep $XYZ
could anybody please tell why this is not working?
Because your $1 variable (first argument to the script) is set to XYZ=KLMN.
Just use abc_sh KLMN and grep $1 instead of grep $XYZ.
(Assuming we are talking about bash here)
The other alternative is defining a temporary environment variable in which case you would have to call it like this: XYZ=KLMN abc_sh
EDIT:
Found what you were using, you have to use set -k (see SHELL BUILTIN COMMANDS in the BASH manual)
-k All arguments in the form of assignment statements are
placed in the environment for a command, not just those
that precede the command name.
So
vinko#parrot:~$ more abc
#!/bin/bash
echo $XYZ
vinko#parrot:~$ set -k
vinko#parrot:~$ ./abc XYZ=KLMN
KLMN
vinko#parrot:~$ set +k
vinko#parrot:~$ ./abc XYZ=KLMN
vinko#parrot:~$
So, the place where this was working probably has set -k in one of the startup scripts (bashrc or profile.)
Try any of these to set a temporary environment variable:
XYZ=KLMN abc_sh
env XYZ=KLMN abc_sh
(export XYZ=KLMN; abc_sh)
you are using so many commands chained together....
ps -u `id -u` -f | awk -v x="$XYZ" -v p="ppLSN" '$0~p{
m=split($9,a,"=")
if(a[2]~x){count++}
}
END{print count}'
Call this script:
#!/bin/ksh
ps -u $(id -u) -o args | grep $XYZ | cut -f2- -d " "
Like this:
XYZ=KLMN abc_sh

Multiple grep search/ignore patterns

I usually use the following pipeline to grep for a particular search string and yet ignore certain other patterns:
grep -Ri 64 src/install/ | grep -v \.svn | grep -v "file"| grep -v "2\.5" | grep -v "2\.6"
Can this be achieved in a succinct manner? I am using GNU grep 2.5.3.
Just pipe your unfiltered output into a single instance of grep and use an extended regexp to declare what you want to ignore:
grep -Ri 64 src/install/ | grep -v -E '(\.svn|file|2\.5|2\.6)'
Edit: To search multiple files maybe try
find ./src/install -type f -print |\
grep -v -E '(\.svn|file|2\.5|2\.6)' | xargs grep -i 64
Edit: Ooh. I forgot to add the simple trick to stop a cringeable use of multiple grep instances, namely
ps -ef | grep something | grep -v grep
Replacing that with
ps -ef | grep "[s]omething"
removes the need of the second grep.
Use the -e option to specify multiple patterns:
grep -Ri 64 src/install/ | grep -v -e '\.svn' -e file -e '2\.5' -e '2\.6'
You might also be interested in the -F flag, which indicates that patterns are fixed strings instead of regular expressions. Now you don't have to escape the dot:
grep -Ri 64 src/install/ | grep -vF -e .svn -e file -e 2.5 -e 2.6
I noticed you were grepping out ".svn". You probably want to skip any directories named ".svn" in your initial recursive grep. If I were you, I would do this instead:
grep -Ri 64 src/install/ --exclude-dir .svn | grep -vF -e file -e 2.5 -e 2.6
you can use awk instead of grep
awk '/64/&&!/(\.svn|file|2\.[56])/' file
You maybe want to use ack-grep which allow to exclude with perl regexp as well and avoid all the VC directories, great for grepping source code.
The following script will remove all files except a list of files:
echo cleanup_all $#
if [[ $# -eq 0 ]]; then
FILES=`find . -type f`
else
EXCLUDE_FILES_EXP="("
for EXCLUDED_FILE in $#
do
EXCLUDE_FILES_EXP="$EXCLUDE_FILES_EXP./$EXCLUDED_FILE|"
done
# strip last char
EXCLUDE_FILES_EXP="${EXCLUDE_FILES_EXP%?}"
EXCLUDE_FILES_EXP="$EXCLUDE_FILES_EXP)"
echo exluded files expression : $EXCLUDE_FILES_EXP
FILES=`find . -type f | egrep -v $EXCLUDE_FILES_EXP`
fi
echo removing $FILES
for FILE in $FILES
do
echo "cleanup: removing file $FILE"
rm $FILE
done

Kill respawing server running on port 3000 created by `node . > /dev/null 2> /dev/null < /dev/null &`

I'm using AWS CodeDeploy in which server running on pm2 dose not work due to explanation given here in troubleShoot documentation.
I followed the documentation and in AfterInstall script used node . > /dev/null 2> /dev/null < /dev/null & to run the node server in the background.
I've tried following ways to kill the server
fuser -k 3000/tcp
lsof -P | grep ':3000' | awk '{print $2}' | xargs kill -9
kill -9 $(lsof -t -i:3000)
but each time a new process respwans with a different PID.
How can I kill this background process and add it to the ApplicationStop script for CodeDeploy?
One of the problems with finding a pid with grep is that the grep pid will also show up as a result and can kill itself before the target, so try;
ps ax | grep node | grep -v grep
if it looks reasonable, review this;
ps ax | grep node | grep -v grep | awk '{print $1}'
then run the kill;
ps ax | grep node | grep -v grep | awk '{print $1}' | xargs kill -9
pkill is a less flexible option (no regex filtering) but if you use that be sure to use the -I flag so you don't kill anything you did not intend to.
I was able to kill using pkill node command.

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