Awk: filter string + count number of lines?

Question

I have some JSON data:

    "Item1": {
      "foo": null,
      "version": "bar",
      "result": null,
    },
    "Item2": {
      "foo": null,
      "version": "bar",
      "result": null,
    },
    "Item3": {
      "foo": null,
      "version": "bar",
      "result": null,
    },

With awk I'm able to filter strings:

$ awk '/version/' /tmp/json
      "version": "bar",
      "version": "bar",
      "version": "bar",

I'm trying to count the number of lines and get the following result without piping, pure awk.

$ awk '/version/' /tmp/json | wc -l
3

Examples online show how to use END and NR but this doesn't produce the results I'm looking for:

$ awk '/version/{print NR}' /tmp/json
3
8
13

or

$ awk 'END/version/{print NR}' /tmp/json
awk: line 1: syntax error at or near /version/

Answer 1

If you really insist on doing this in awk, all you need to do is increment a variable for each matching line and then, once all lines have been processed, print that variable:

$ awk '/version/{c++}END{print c}' file
3

Or, if you want to get a 0 printed when there are no matches (the above will just print an empty line), use:

awk '/version/{c++}END{print c+0}' file

Finally to print only if there is at least one match (so no output, instead of an empty line, when there are no matches), use:

awk '/version/{c++}END{ if(c) print c}' 

Answer 2

awk 'BEGIN { count=0 } /version/ { count++ } END { printf( "%d\n", count ) }' datafile.json

But, more simply:

grep -c 'version' datafile.json