Bash: create a report with grep output?

Greetings.

I need to generate a simple report via Bash (Korn?) with this raw data

Test_Version=V2.5.2
Test_Version=V2.6.3
Test_Version=V2.4.7
Test_Version=V2.5.2
Test_Version=V2.5.2
Test_Version=V2.5.1
Test_Version=V2.5.0
Test_Version=V2.3.9
Test_Version=V2.3.1

Ideally, I'd like to get something like this sorted output

Version    Count
...
V2.5.0     1
V2.5.1     1
V2.5.2     3
V2.6.3     1
...

I can sort the output like this (raw data is contained in ASCII files):

find . -name "*.VER" -exec grep "Test_Version" '{}' ';' -print | grep -e "Test_Version" | sort -u

But I can't figure out how to count my rec开发者_开发百科ords in a tabular layout. Any idea how could I do that?

Thanks!!

What about something like:

$ cat input.txt | sed 's/.*=//' | sort | uniq -c
      1 V2.3.1
      1 V2.3.9
      1 V2.4.7
      1 V2.5.0
      1 V2.5.1
      3 V2.5.2
      1 V2.6.3

Can tweak it into the exact format from there...

This seems like a job for awk:

Assuming your version information is in the file versions.txt (you can also not specify a filename, in that case awk reads from stdin).

awk -F= '
        {counts[$2]=counts[$2]+1}
END     {for (key in counts)
            printf "%s\t%d\n", key, counts[key]}
' versions.txt

Explanation:

• -F= tells awk to use the = character as field separator. Each line in your data will be treated as two fields of which only the second is used.
• The first statement between braces is executed for each line of input. Keeping count for each occurence of the second field, which is $2.
• The second statement in braces preceded by the keyword END is executed after the last line is processed. It shows all the counts for all distinct values of $2.

You won't need any greps:

find . -name "*.VER" | 
  awk -F= '
    BEGIN {
      OFS="\t"
    }
    /Test_Version/{
      if (!count[$2]++) ver[num++]=$2
    } 
    END {
      print "Version", "Count"
      n=asort(ver); 
      for (i=1;i<=n;i++) print ver[i], count[ver[i]]
    }'