When executing ./test.sh 12.34, the grep should match 12.34 and not 12-34. How can this be accomplished?
#!/bin/sh
ip=$1
echo $ip
if netstat | grep ssh | grep $ip; then
netstat | grep ssh | grep $ip
else
echo 'No'
fi
You could use grep with the -F option:
From man grep:
-F, --fixed-strings
Interpret pattern as a set of fixed strings (i.e. force grep to
behave as fgrep).
Your example:
grep -F "$ip"
grep using regex to match strings. . is a special character in regex, so it needs to be escaped. There is a rather elegant way of doing this:
export escaped_ip_addr = $(echo $ip_addr | sed "s/\./\\\./g")
Which would make your final code:
#!/bin/sh
#test.sh
ip=$1
echo $ip
export escaped_ip = $(echo $ip | sed "s/\./\\\./g")
if netstat | grep ssh | grep $escaped_ip; then
netstat | grep ssh | grep $escaped_ip
else
echo 'No'
fi
Related
I have a string,
var=refs/heads/testing/branch
I want to get rid of refs/heads/ in the string using shell script, such that I have only:
var=testing/branch
Commands I tried (one per line):
echo $(var) | awk -F\\ {'print $2'}
echo $var | sed -e s,refs/heads/,,
echo "refs/heads/testing/branch" | grep -oP '(?<=refs/heads/\)\w+'
echo "refs/heads/testing/branch" | LC_ALL=C sed -e 's/.*\\//'
echo "refs/heads/testing/branch" | cut -d'\' -f2
echo refs/heads/testing/branch | sed -e s,refs/heads/,,
there are lots of options out there ,try easy ones:
echo $var | cut -d "/" -f 3,4
echo $var | awk -F"/" '{print $3"/"$4}'
Shell parameter expansion: remove the prefix "refs/heads/" from the variable contents
$ var=refs/heads/testing/branch
$ echo "${var#refs/heads/}"
testing/branch
so i made a bash script the greps the name of the host on the redirected file. However, there are hosts that are named either with "-" or "_"
GTR_SRV123_EST
GTR-SRV123-EST
Right now, what i did was, grep just a portion of the FQDN, like SRV123
Is there a way i can grep the host even if i just put the FQDN GTR_SRV123_EST and it will still matched this GTR-SRV123-EST.
i have a prompt that ask for the hostname:
echo -n "Please enter the host: "
read $host
grep -i $host ${temp}/*
update:
so had it working with the help of Juan's command. However, the directory path is displayed on the output. How can i get rid of it.
/export/home/aa12s/GLB-TXU/temp/
Current output:
/export/home/aa12s/GLB-TXU/temp/GBL-ASA-A:100022FBC0D00038 gbl-asa-a-mode1 5005076801103673 active gbl-ac-wbg02
Desired output:
GBL-ASA-A:100022FBC0D00038 gbl-asa-a-mode1 5005076801103673 active gbl-ac-wbg02
Command:
grep -iE "$(echo $host| awk -F '/export/home/aa12s/GLB-TXU/temp/' '{$2=$1;a=gsub(/_/, "-",$2); print $1"|"$2}' 2>/dev/null)" ${temp}/*
Edit your pattern.
echo -n "Please enter the host: "
read host # Edit: not $host
host="${host//[_-]/\[_-\]}" # turn either into a *check* for either
grep -i "$host" ${temp}/*
Kind of hacky but give this a try:
grep -Ei "$(echo $host| awk '{$2=$1;gsub(/_/, "-",$2);print $1"|"$2}' 2>/dev/null)" ${temp}/*
To get rid of filepaths:
grep -iE "$(echo $host| awk '{$2=$1;gsub(/_/, "-",$2);print $1"|"$2}' 2>/dev/null)" ${temp}/* 2>/dev/null|awk -F \/ '{print $NF}'
NOTE: Slashes must NOT be present in the file content.
If there is no host name with both _ and - below will work.
Entered host contains only _
grep -iE $(echo $host | tr "_" "-")\|$host ${temp}/*
Entered host contains only -
grep -iE $(echo $host | tr "-" "_")\|$host ${temp}/*
Entered host contains both _ and -
grep -iE $(echo $host | tr "_" "-")\|$(echo $host | tr "-" "_")\|$host ${temp}/*
You can use backreference :
([_-]) : capture either _ ou - in group 1
\1 : reference group 1
try this command :
grep -iE "([_-])$host\1" ${temp}/*
https://regex101.com/r/uH5SHC/1/
Wiyh host=SRV123, you will capture :
GTR_SRV123_EST
GTR-SRV123-EST
and not
GTR_SRV123-EST
So I have an expression that I want to extract some lines from a text and count them. I can grep them as follows:
$ cat medsCounts_totals.csv | grep -E 'NumMeds": 0' | wc -l
Which is fine. Now I want to loop over with the string ...
$ for i in {0..10}; do expr="NumMeds\": $i"; echo $expr; done
However, when I try to use $expr
for i in {0..10}; do expr="NumMeds:\" $i"; cat medsCounts_totals.csv | grep -E "$expr" | wc -l ; done
I get nothing. How do I solve this problem in an elegant manner?
there is a typo in
for i in {0..10}; do expr="NumMeds:\" $i"; cat medsCounts_totals.csv | grep -E "$expr" | wc -l ; done
it should be
"NumMeds\": $i"
The script:
#!/bin/bash
SERVICE="$1"
RESULT=`ps -a | sed -n /${SERVICE}/p`
MEM=$(ps aux | sort -rk +4 | grep $1 | grep -v grep | awk '{print $4}' | awk 'NR == 1')
if [ "${RESULT:-null}" = null ]; then
echo "$1 is NOT running"
else
echo "$MEM"
fi
if [ "$MEM" -ge 1 ]; then
mailx -s "Alert: server needs to be checked" me#admins.com
fi
The problem with the ps output piped to sed is even if no process is running it finds itself:
ps aux | sed -n /nfdump/p
root 30724 0.0 0.0 105252 884 pts/1 S+ 14:16 0:00 sed -n /process/p
and them my script bypasses the expected result of "service is not running" and goes straight to echo'ing $MEM, which in this case will be 0.0. Does sed have a grep -v grep eqivalent to get itself out of the way?
Let me add one more example in addition to my comments above. Sed has indeed an equivalent to grep -v: You can negate a match with /RE/! (append a ! to the regex) thus you can try:
ps aux | sed -n '/sed/!{ /nfdump/ p;}'
Here the part inside { ... } is only applied to lines not matching sed.
For the record: there is a command pgrep that can replace your ps sed pipeline, see pgrep in Wikipedia or its manpage.
Try :
ps aux | sed -n '/[n]fdump/p'
or :
ps aux | grep '[n]fdump'
The regex won't be find in the processes list
I get the ip address like that :
Ip=`ifconfig | grep inet | grep -v -E 'inet6|127.0.0.1' | \
tr -d [:alpha:] | tr -s [:space:] | cut -d: -f2`
I have an ip like this for instance : 10.1.0.76
I want to make a new variable with the Ip variable to have another ip, for instance my new variable will return : 10.1.0.178
Just the last number change, so I want to get just a part of Ip variable (10.1.0.) and add another number to the end.
I tried with sed but I always have mistakes like "there's no file call'd ..."
Can you help me ?
You can use parameter expansion: It's simply: ${Ip%.*}.178
${Ip%.*} is the ip with the last dot and everything after it removed. The .178 is what you want to append after that.
Here it is in context:
# Your original expression
Ip=`ifconfig | grep inet | grep -v -E 'inet6|127.0.0.1' | \
tr -d [:alpha:] | tr -s [:space:] | cut -d: -f2`
# assign a new variable with the ip with different end octet
newIp=${Ip%.*}.178
# Show new ip
echo "$newIp"
Well, given that you have IP in a format x.y.z.w, you can use perl regex:
$ echo "120.20.31.78" | perl -pe 's/(.*)\..*/$1\.123/'
120.20.31.123
This will repace last number ("78") with "123".
So, in your case (assuming your "Ip" variable is set correctly), it would be:
Ip=ifconfig | grep inet | grep -v -E 'inet6|127.0.0.1' | tr -d [:alpha:] | tr -s [:space:] | cut -d: -f2 | perl -pe 's/(.*)\..*/$1\.123/'
see this, I hope it is what you want:
kent$ echo $ip
10.1.0.76
kent$ echo $part
178
kent$ sed -r "s/(.*\.).*/\1$part/" <<< $ip
10.1.0.178
to set $ip with new value:
kent$ ip=$(sed -r "s/(.*\.).*/\1$part/" <<< $ip)
kent$ echo $ip
10.1.0.178