Bash get substring to variable using regex

Question

Using bash script I need to search string like version = '1.8.1-SNAPSHOT' from text file and get value '1.8.1-SNAPSHOT' into variable.

I tryed to use the next code, but no result:

version=$(grep -P "version\s?=\s?'([a-zA-Z.\d-]){5,20}?'" file.txt) 
regex="'([a-zA-Z.\d-]){5,30}?'"
value=`expr match "$version" '([a-zA-Z.\d-]){5,30}?'`

What is wrong and are there another way?

Some of text file:

buildscript {
    repositories {
        mavenCentral()
    }
    dependencies {
        classpath('org.springframework.boot:spring-boot-gradle-plugin:1.5.2.RELEASE')
    }
}

springBootVersion = '2.0.7.RELEASE'


apply plugin: 'java'
apply plugin: 'idea'
apply plugin: 'org.springframework.boot'

// for Glassfish
//apply plugin: 'war'

jar {
    baseName = 'work-space'
    version = '1.8.1-SNAPSHOT'
}

Desired string need to be '1.8.1-SNAPSHOT' into variable for the next manipulation.

Try version="$(grep -oP "version\s*=\s*'\K[^']+" file)". See ideone.com/LtxK2g — Wiktor Stribiżew
– Wiktor Stribiżew, Commented Dec 16, 2018 at 10:09
Thanks, but i also need to get from version = '1.8.1-SNAPSHOT' the finish result like '1.8.1-SNAPSHOT' — jlemon
– jlemon, Commented Dec 16, 2018 at 10:11
ideone.com/LtxK2g shows the 1.8.1-SNAPSHOT - isn't it expected? — Wiktor Stribiżew
– Wiktor Stribiżew, Commented Dec 16, 2018 at 10:13

Wiktor Stribiżew · Accepted Answer · 2018-12-16 10:42:48Z

4

You may use

version="$(grep -oP "version\s*=\s*'\K[^']+" file)"

See the online demo.

As you are using P option, I assume you may use PCRE regex with your grep. To output the match, you also need to add o option.

The regex you need is version\s*=\s*'\K[^']+:

version - matches version substring
\s*=\s* - = enclosed with 0+ whitespaces
' - a ' char
\K - match reset operator discarding all text matched so far
[^']+ - 1 or more chars other than '

answered Dec 16, 2018 at 10:42

Wiktor Stribiżew

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