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Bash: create a report with grep output?

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# 1  
Old 12-30-2010
Bash: create a report with grep output?
Greetings.

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

Code:
    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

Code:
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):

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

But I can't figure out how to count my records in a tabular layout. Any idea how could I do that?

Thanks!!
# 2  
Old 12-30-2010
Try: sort | uniq -c instead of sort -u
# 3  
Old 12-30-2010
Thanks for your reply. This is what I got

1)
Code:
find . -name "*.VER" -exec grep "Test_Version" '{}' ';' -print | grep -e "Test_Version" > outfile

The output of 1) looks like this

Code:
...
Test_Version=2.5.2
Test_Version=2.6.0
Test_Version=2.6.2
...

2)
Code:
cat outfile | sed 's/.*=//' | sort | uniq -c

My output looks like this:
Code:
     39 V2.0.0
   3220 V2.4.7
   7967 V2.5.0
  28463 V2.5.2

Thanks to a friend who's way more clever than me.Smilie

Al.
# 4  
Old 12-30-2010
by awk without temp file generated.

Code:
awk -F= '/Test_Version/ {a[$2]++}END{for (i in a) print a[i],i|"sort -n"}' *.VER

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