How to use grep to capture match?

Question

I'm trying to write a script that increments an integer found in a config file. I'm trying to use grep to find the current value:

$ grep versionNumber myfile.conf
          versionNumber 123789
^ whitespace
          ^ the key I need to match
                        ^ the value I need to capture

My desired result is to capture the 123789 in the example above.

What do I need to do in order to capture this value?

Answer 1

You can do this:

grep versionNumber myfile.conf | grep -oE '[0-9]+'

or this

grep versionNumber myfile.conf | awk '{print $2}'

or this

grep versionNumber myfile.conf | cut -d' ' -f2

or if you have GNU grep with support for the -P mode:

grep -oP 'versionNumber \K\d+' myfile.conf

Answer 2

使用awk

awk '/versionNumber/{print $2}' myfile.conf

Answer 3

Using grep -oP (pcre mode);

s='          versionNumber 123789'
grep -oP 'versionNumber\h+\K\S+' <<< "$s"

123789

Answer 4

script that increments an integer found in a config file using Gnu awk v. 4.1.0+:

A test file:

$ cat > file
foo
           versionNumber 123789
bar

Using inplace edit feature:

$ awk -i inplace '/versionNumber/ {n=index($0,"v"); sub($2,$2+1); printf "%-"n"s\n", $0; next} 1' file

The result:

$ cat file
foo
           versionNumber 123790
bar

Explanation:

Key to editing the file inplace is -i inplace

/versionNumber/ {          # if versionNumber found in record
    n=index($0,"v")        # count leading space amount to n
    sub($2,$2+1)           # increment the number in second field
    printf "%-"n"s\n", $0  # print n space before the (trimmed) record
    next                   # skip print in the next line
} 1                        # print all other records as were