bash + regular expression + capture java version [closed]

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how to capture the jdk1.8.0_112 under /usr/jdk64

 /usr/jdk64/jdk1.8.0_112/

while jdk1.8.0_112 is example

and could be any other version as

jdk1.7.0_112
jdk3.8.2_112
jdk1.8.0_117
jdk1.2.0_112
jdk2.7.66_112
jdk1.8.2_111
jdk1.23.0_117
jdk9.2.093_112

Unclear: What do you mean by "capture"? Is the list that you show names of subdirectories that exists under /usr/jdk64? What is wrong with expanding the shell glob /usr/jdk64/jdk*? What are you expecting as the result (a string in a variable?) — Kusalananda, Commented May 22, 2019 at 9:49
What’s the higher-level problem you’re trying to solve? Why do you need this particular string, and what are you planning on doing with it? — Stephen Kitt, Commented May 22, 2019 at 10:06

jesse_b · Accepted Answer · 2019-05-22 14:03:31Z

1

find /usr/jdk64/* -prune -type d -name 'jdk*' | grep -P 'jdk\d+\.\d+\.\d+_\d+$'

jesse_b

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answered May 22, 2019 at 9:41

dedowsdi

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Juan · Accepted Answer · 2019-05-22 09:42:39Z

0

One possibility for the regexp you need is:

grep -E "jdk([0-9]+).([0-9]+).([0-9_]+)"

answered May 22, 2019 at 9:42

Juan

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2 Answers 2