Could not extract jdk-version using grep-command and regular expressions - regex

Question

I am trying to extract the version of a jdk using regular-expressions Actually I have the following version:
openjdk 11.0.9.1 2020-11-04
OpenJDK Runtime Environment (build 11.0.9.1+1-Ubuntu-0ubuntu1.18.04)
OpenJDK 64-Bit Server VM (build 11.0.9.1+1-Ubuntu-0ubuntu1.18.04, mixed mode,sharing)
I wrote the regex. It looks like the following:
^[0-9]{1,2}\.[0-9]{1,2}\.[0-9]{1,2}\.[0-9]{1,2}$
This regular expression is working on the online regex tester (https://regex101.com/). Unfortunately, it is not working with grep-command. I am using the extended regular expression. My Code for the extraction of jdk-version look like the following:
CMD_RESULT=$(java --version 2>&1 | head -n 1 | cut -d '' -f 2)
if [ ! -z "$CMD_RESULT" ]
then
for token in $CMD_RESULT
do
JAVA_VERSION=$(echo $token | grep -e $VERSION_EXTRACTION_REGEX)
if [ ! -z $JAVA_VERSION ];
then
printf "${GREEN}Java version: [$JAVA_VERSION]\n"
fi
done
fi
I am not understanding, why my regex is working on the online regex checker, while it is not working with the grep-command. I am using the Parameter -e for extended regexp.
Any Idea ?

Answer 1

A few small changes to OPs current code:
as mentioned in the comments, ^ and $ represent the beginning and ending of the input; remove these from the regex
while grep can use the resulting regex it will be necessary to tell grep to run in -Extended regex mode
we can use grep's -o option to limit output to just the portion that matches the regex
Sample input data:
$ cat jdk.dat
openjdk 11.0.9.1 2020-11-04
OpenJDK Runtime Environment (build 11.0.9.1+1-Ubuntu-0ubuntu1.18.04)
OpenJDK 64-Bit Server VM (build 11.0.9.1+1-Ubuntu-0ubuntu1.18.04, mixed mode,sharing)
One grep solution using OP's modified regex:
$ grep -oE '[0-9]{1,2}\.[0-9]{1,2}\.[0-9]{1,2}\.[0-9]{1,2}' jdk.dat
11.0.9.1
11.0.9.1
11.0.9.1
Same thing but with the regex stored in OPs variable:
$ VERSION_EXTRACTION_REGEX='[0-9]{1,2}\.[0-9]{1,2}\.[0-9]{1,2}\.[0-9]{1,2}'
$ grep -oE "${VERSION_EXTRACTION_REGEX}" jdk.dat
11.0.9.1
11.0.9.1
11.0.9.1
NOTE: I'll leave it up to the user to decide which value to use, eg, head -n 1 to work with just the first input line (openjdk 11.0.9.1 2020-11-04)

Answer 2

Be done with using only native Bash commands to parse the java version string:
#!/usr/bin/env bash
# Capture java command path if it exists OR exit fail
java_cmd=$(command -v java) || exit 1
# Capture java raw version output
java_raw_version=$("$java_cmd" -version 2>&1)
# Match java raw version string against Bash Regex OR dump debug if match fail
[[ $java_raw_version =~ ([0-9]+)\.([0-9]+)\.([0-9]+)[_.+-]([0-9]+) ]] ||
typeset -p java_raw_version BASH_REMATCH
major=${BASH_REMATCH[1]} # Regex capture group 1
minor=${BASH_REMATCH[2]} # Regex capture group 2
patch=${BASH_REMATCH[3]} # Regex capture group 3
release=${BASH_REMATCH[4]} # Regex capture group 4
# Recompose a semantic version string
semver="$major.$minor.$patch-$release"
printf 'Java semantic version is: %s\n' "$semver"
Sample outputs:
From:
openjdk version "1.8.0_275"
OpenJDK Runtime Environment (build 1.8.0_275-8u275-b01-0ubuntu1~20.10-b01)
OpenJDK 64-Bit Server VM (build 25.275-b01, mixed mode)
Output:
Java semantic version is: 1.8.0-275
From:
openjdk 11.0.9.1 2020-11-04
OpenJDK Runtime Environment (build 11.0.9.1+1-Ubuntu-0ubuntu1.18.04)
OpenJDK 64-Bit Server VM (build 11.0.9.1+1-Ubuntu-0ubuntu1.18.04, mixed mode,sharing)
Output:
Java semantic version is: 11.0.9-1
About semantic versioning, see: https://semver.org/

Answer 3

Here is a very simple awk solution (no need for regexp):
#!/usr/bin/env bash
# Capture java command path if it exists OR exit fail
java_cmd=$(command -v java) || exit 1
# Capture java raw version output
java_raw_version=$("$java_cmd" -version 2>&1)
# print out the Java version:
echo $java_raw_version | awk -F \" 'NR==1{print "Java semantic version is: "$2}''
Explanation:
-F \" set awk fields separator to "
NR==1 only deal with line #1
print "Java semantic version is: "$2 append 2nd field to text and print