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Shell Script to count different extention of each file

Our application we have lot of files.
each file has it's  3-different extention.

for example.....

testfile.ext1
testfile.ext2
testfile.ext2

report.ext1
report.ext2



I want to have shell script, which can test, how many extentions had for each file.
so that, I can easily track missting files.

In otherwords .... , from the above example of files which were listed, I want to have out put like below.

OutPut
==================
testfile ----- has count ---- 3
report  ----- has count -----2

please advice
0
mac_g
Asked:
mac_g
  • 2
4 Solutions
 
arnoldCommented:
The simplest is to use perl
Presuming you are looking at the current level
As well as there is only one period in the filename and

ls | perl -e 'while (<>) {
chomp();
next if test -d "$_";
($filename,$extension)=split($_,2);
$list{$filename}+=1;
}
foreach $extension (keys %list) {
print "$extension --------> $list{$extension}\n";
}
'

Remove the last characters if it is a Control character(cr or lf)
skip directory names in the listing.
Build a hash with the name portion of the file.
0
 
simon3270Commented:
Or using standard tools:
$ ls
double.ext1  mine.ext1	  report.ext2	 testfile.ext2
double.ext2  report.ext1  testfile.ext1  testfile.ext3
$ ls | sed 's/\.[^.]*$//'| sort | uniq -c
      2 double
      1 mine
      2 report
      3 testfile
$

Open in new window

0
 
serialbandCommented:
Or with awk instead of sed:

ls |awk -F"." '{print $1}'|uniq -c

The extra sort in the sed example may not be necessary with a standard presorted ls.
0
 
arnoldCommented:
simon3270 in the sed pattern match took care of having a period in the name
user.lastname.ext which I "excluded as a posibility.

The issue that is not clear is whether the user is looking through a directory hierarchy
such that you can have
file.txt
directory1\file.txt

i.e. user A always copies and saves files into directory1.

Yes, I realize that similarly named files are not necessarily the same file with the same content.
0

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