Get String between 2 Strings or delimeters in bash using regex

Question

My String is:

model_config_list: {
   config: {
       name: "123-model",
       base_path: "/modelServers/tests-integration/resources/models/123-model",
       model_platform: "tensorflow",
       model_version_policy: {
           all: {}
       }
   }

I want to extract:

/modelServers/tests-integration/resources/models/123-model

for that I wrote bash script :

config_file=$(<conf.conf)
regex="(?<=base_path:)(.*)(?=,)"
if [[ $config_file =~ $regex ]];then echo ${BASH_REMATCH[1]}; fi

However, I am not getting any output. I am a beginner with Bash scripting.

Wiktor Stribiżew · Accepted Answer · 2018-05-31 10:50:18Z

3

You need to use a regex with a capturing group rather than lookarounds that are not supported in Bash regex:

regex='base_path:[[:space:]]*"([^"]+)"'
if [[ $config_file =~ $regex ]];then 
  echo ${BASH_REMATCH[1]};
fi

Output:

/modelServers/tests-integration/resources/models/123-model

See the Bash demo online.

Details

The positive (?<=base_path:) lookbehind and (?=,) lookahead are not supported in Bash regex. The point here is to match base_path: with any whitespace after it, then ", and then capture into Group 1 any one or more chars other than " (using a negated bracket expression [^"]+).

base_path: - a literal substring
[[:space:]]* - 0+ whitespace chars
" - a double quote
([^"]+) - 1 or more chars other than a " char
" - a double quote.

answered May 31, 2018 at 10:50

Wiktor Stribiżew

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Comments

choroba · Accepted Answer · 2018-05-31 10:51:10Z

2

Bash doesn't support lookaround assertions. But you don't need them:

regex='base_path: "([^"]*)"'

answered May 31, 2018 at 10:51

choroba

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