Regex for pattern matching in shell script and extract the match part

Question

Hope you are doing.

I need help in regex in the shell script.

the input is:

PM_path='SubNetwork=IMS,SubNetwork=IMS,MeContext=R9NEXRERNFVCSS01_PS'**PM_path='SubNetwork=IMS,SubNetwork=IMS,MeContext=R9NEXRERNFVCSS01_PS'

the output I want as:

R9NEXRERNFVCSS01_PS that is everything after the last equal sign.

Right now I have implemented it as below:

if [[ $PM_path =~ MeContext=([a-zA-Z0-9_]+) ]]; then  
      NODE_NAME=${BASH_REMATCH[1]}
 echo "the value is matched";

So here I have put a check at MeContext and this is a part of PM_path.

I wanted to make it more generic like select everything which appears after the last equal sign.

Please help.

Thanks in advance.

RavinderSingh13 · Accepted Answer · 2021-05-08 08:45:38Z

6

You could use grep here, with GNU grep please try following. Simply using grep's -o and -P option(to enable PCRE regex). Then in regex matching everything till last occurrence of = and using \K to discard matched part and .* followed by it will print the rest of the value from variable.

echo "$PM_path" | grep -oP '.*=\K.*'

answered May 8, 2021 at 8:45

RavinderSingh13

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Comments

anubhava · Accepted Answer · 2021-05-10 13:32:12Z

4

I wanted to make it more generic like select everything which appears after the last equal sign.

You may use:

[[ $PM_path =~ .*=([^/]+) ]] && echo "${BASH_REMATCH[1]}"
R9NEXRERNFVCSS01_PS

.* matches longest possible text from start then we match a =. Finally we match and capture remaining string of 1+ non-/ characters in ([^/]+) that we print using echo "${BASH_REMATCH[1]}"

edited May 10, 2021 at 13:32

answered May 8, 2021 at 13:49

anubhava

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2 Comments

Chanchal singh Over a year ago

Hi Anubhava, Hope you are doing good. I have one more issue. The exact PM_path is "PM_Path=**"SubNetwork=IMS,SubNetwork=IMS,MeContext=R9NEXRERNFVCSS01_PS/"** so after ${BASH_REMATCH[1]} the output is as R9NEXRERNFVCSS01_PS/ i.e with extra slash. Ccould you please help me out to get rid of slash..

anubhava Over a year ago

@Chanchalsingh: Please check my updated answer now

Wiktor Stribiżew · Accepted Answer · 2021-05-11 08:34:26Z

You can simply get the value using Bash string manipulation:

PM_path="${PM_path##*=}"

Here, ${PM_path##*=} means that you want to remove as many chars as possible from the left (##*) till the last = char.

See the 10.1. Manipulating Strings reference:

${string##substring}
Deletes longest match of $substring from front of $string.

See an online demo:

PM_path='SubNetwork=IMS,SubNetwork=IMS,MeContext=R9NEXRERNFVCSS01_PS'
PM_path="${PM_path##*=}"
echo "$PM_path"
# => R9NEXRERNFVCSS01_PS

ANSWER TO YOUR COMMENT:

You mention you have a trailing / char at the end of your string. You can still use the same technique, remove / from the result obtained with the preceding step:

PM_Path="SubNetwork=IMS,SubNetwork=IMS,MeContext=R9NEXRERNFVCSS01_PS/"
PM_Path="${PM_Path##*=}"
PM_Path="${PM_Path/\//}"
echo "$PM_Path"

See the online demo, PM_Path="${PM_Path/\//}" removes the first / in the string.

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