bash: find all regex in file and put into array

Question

I have a template config file with PlaceHolders, and i want to find all those PlaceHolders and put inside an array.

The curent state:
I can find all the PlaceHolders in file only if there is no more than one PlaceHolder in a line.

For example, this is my template file:

upstream portal {        
   server {{UNICORN_SERVICE_NAME}}:{{UNICORN_SERVICE_PORT}};  
}

server {
  listen *:80 default_server;         
  server_name {{APP_HOST}};     
  server_tokens off;     
  root /dev/null;

  # Increase this if you want to upload large attachments
  # Or if you want to accept large git objects over http
  client_max_body_size {{NGINX_MAX_UPLOAD_SIZE}};

  location {{GITLAB_RELATIVE_URL_ROOT}}/ {
    root /var/lib/nginx/portal;
    # serve static files from defined root folder;.
    # @gitlab is a named location for the upstream fallback, see below    
  }
  }

this is the code i use to find the PlaceHolders:

matches_bloc=$(awk 'match($0, /(\{\{)([^{]*)(\}\})/) {
                    print substr($0, RSTART, RLENGTH)                    
                }' ${currFile})

            # convert 'matches_bloc' into array
            matches=()
            echo "Matches:"
            while read -r line; do
                matches+=("$line")
                echo "  - ${line}"
            done <<< "$matches_bloc"

in this example the matches result will be:

Matches:
- {{UNICORN_SERVICE_NAME}}
- {{APP_HOST}}
- {{NGINX_MAX_UPLOAD_SIZE}}
- {{GITLAB_RELATIVE_URL_ROOT}}

You can notice that there are 5 PlaceHolders in the file and only 4 matches.
The missing match is: {{UNICORN_SERVICE_PORT}}, because there is already another match in the same line.

My question is:
How can i find all the matches in the file regardless of the line?

Community · Accepted Answer · 2020-06-20 09:12:55Z

6

Find all variables in a template file and put them in an array.

With GNU grep:

array=( $(grep -Po '{{.*?}}' file) )
declare -p array

Output:

declare -a array='([0]="{{UNICORN_SERVICE_NAME}}" [1]="{{UNICORN_SERVICE_PORT}}" [2]="{{APP_HOST}}" [3]="{{NGINX_MAX_UPLOAD_SIZE}}" [4]="{{GITLAB_RELATIVE_URL_ROOT}}")'

-P: Interpret {{.*?}} as a Perl regular expression.

-o: Print only the matched (non-empty) parts of a matching line, with each such part on a separate output line.

* The preceding expression can match zero or more times. With ? the * tries to match as few as possible (non-greedy).

See: The Stack Overflow Regular Expressions FAQ

edited Jun 20, 2020 at 9:12

CommunityBot

11 silver badge

answered Jun 4, 2017 at 13:41

Cyrus

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2 Comments

RavinderSingh13 Over a year ago

Thank you cyrus for great code, could you please explain the regex part, will be grateful to you sir.

Sundeep Over a year ago

can be done without -P as well... grep -o '{{[^}]*}}' file

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