Very Computer

Good day,

I need to write a shellscript and I hope you can help me:

I have got 30 directories, each has a different name. In the
directories, files are stored which are built up like this: XXX.000,
XXX.001 ....up to XXX.500

XXX can be everything from i.e. test1(.066) to actock77(.003).

Now I need to find the file with the highest extension (i.e. XXX.050)
and delete all files that have a lower extension (XXX.000 - XXX.049),
so that only XXX.050 remains. Approximately 2300 files are to be
deleted.

After that, the computer should switch to the next directory and
continue the progress.

With kindest regards,

Chris

[...]

Quote:> I have got 30 directories, each has a different name. In the
> directories, files are stored which are built up like this: XXX.000,
> XXX.001 ....up to XXX.500

> XXX can be everything from i.e. test1(.066) to actock77(.003).

> Now I need to find the file with the highest extension (i.e. XXX.050)
> and delete all files that have a lower extension (XXX.000 - XXX.049),
> so that only XXX.050 remains. Approximately 2300 files are to be
> deleted.

> After that, the computer should switch to the next directory and
> continue the progress.

With the Z-shell.

zmodload zsh/files
dirs=( dir1 dir2 dir3 ... )
for d in $dirs; do
  files=( $d/*.{000..500}(N) )
  rm -f -- $files[1,-2]
done

Note that only one file with the highest extension will remain.

--
Stphane

One way to list the files in order is ls -r XXX.*
and we do not want the highest one so pipe to sed -n '2,$p'
and then we want to remove them: pipe to xargs -r rm
(Gnu xargs has the -r option which means do nothing if
there are no files -- if your xargs does not have this then
you need to work out what should happen if there are
no files. Or install Gnu xargs. Or create "dummy" files first.
Or count how many files there are first.)

So, assuming we know the directories in advance (say, A, B, C):

for dir in A B C
do
    ls -r i${dir}/XXX.* | sed -n '2,$p'
done | xargs -r rm

Or you may decide it is clearer if the xargs is inside the loop:

for dir in A B C
do
    ls -r i${dir}/XXX.* | sed -n '2,$p'  | xargs -r rm
done

if we do not know the directories in advance, then we can do:
for dir in `find topleveldir -type d`
...

John.

#!/usr/bin/perl -w
use strict;

    'dir1',
    'dir2',
    # ...
    'dir29',
    'dir30',
    );

    # sort files based on three digit extension

                sort { $a->[1] <=> $b[1] }
                map  { [ $_, /\.(\d\d\d)$/ ] }
                grep { /.+\.\d\d\d$/ }
                <$dir/*>;
    # remove highest numbered file from array

    # delete remaining files

        unlink $file or warn "Cannot delete $file: $!";
        }
    }

__END__

John
--
use Perl;
program
fulfillment

Quote:> Good day,

> I need to write a shellscript and I hope you can help me:

> I have got 30 directories, each has a different name. In the
> directories, files are stored which are built up like this: XXX.000,
> XXX.001 ....up to XXX.500

> XXX can be everything from i.e. test1(.066) to actock77(.003).

> Now I need to find the file with the highest extension (i.e. XXX.050)
> and delete all files that have a lower extension (XXX.000 - XXX.049),
> so that only XXX.050 remains. Approximately 2300 files are to be
> deleted.

> After that, the computer should switch to the next directory and
> continue the progress.

> With kindest regards,

> Chris

#! /usr/bin/ksh

dir=.
left=""

ls -R $dir | sort -rn | grep -v ":$" | sed '/^$/d' | tr '.' ' ' |
while read left right ; do
   if [ "$left" = "$leftPrev" -a -n "$right" ] ; then
      # rm -f "$left.$right"
      echo "$left.$right"
   fi
   leftPrev=$left
done

### END OF SCRIPT

Note that this script assumes ALL files under the starting directory
have the XXX.000 format, and that each filename has only one period.

-Bill

[...]

Some points are not clear in the description of your problem:
is the XXX unique in each directory. I mean

in dir1
is there foo.001 bar.002, actock77.003 ...
(and you only want to keep actock77.003)

or just
foo.001, foo.002, foo.003
(and you only want to keep foo.003)

or
foo.001, bar.001, foo.002 bar.002
(and you only want to keep foo.002 and bar.002)

Can the file names contain blanks, quotes or newline characters?

Do the directories contain other kind of files (foo.1, foo.501,
README.txt, sub-directories...)

Are there more than one file with the same extension?

Do you need to descend into subdirectories?

Several solutions have been given, they don't all answer the
same question depending on what the poster understood of your
request.

--
Stphane

#! /usr/bin/ksh
## REVISED

dir=.
left=""

ls -R $dir | sort -rn | grep -v ":$" | sed '/^$/d' | tr '.' ' ' |
while read left right ; do
   if [ "$left" = "$leftPrev" -a -n "$right" ] ; then
      # find $dir -type f -name "$left.$right" | xargs rm -f
      echo "$left.$right"
   fi
   leftPrev=$left
done

Untested, as basic idea

Dirlist='/path1 /path2 ...'
for dir in $Dirlist
do
   cd $dir
   ls *.[0-9][0-9][0-9] | sort -t. -r -k2 | tail +2 | xargs rm
done

Regards
Juergen

will only work if there is a single prefix  XXX
in each directory.

if not, first you have to identify possible different prefices before
your file extensions .nnn

hint: awk and associative arrays

Regards
Juergen

conditions
1. Filenamestructure: what_ever_withoutdot.nnn, n is a digit
2. no subdirectories in /path1, /path2 ...

Dirlist='/path1 /path2 ...' # adjust
Ext='[0-9][0-9][0-9]'
for dir in $Dirlist
do
   cd $dir
   praeflist=`ls *.$Ext 2>/dev/null |cut -d. -f1 | uniq`
   [ -n "$praeflist" ] && for praef in $praeflist
   do
      ls "$praef".$Ext | sort -t. -r -k2 |tail +2|xargs rm -f
   done
done

Regards
Juergen

[...]

Quote:> 1. Filenamestructure: what_ever_withoutdot.nnn, n is a digit

                                            nor_space_nor_single
_or_double_quotes_nor_LF_nor_TAB_nor_backslash ;)

--
Stphane

1. Determine 'XXX'.  Find the highest number for each 'XXX', and delete any
lower numbers:

    ls *.[0-9][0-9][0-9] | sed 's/\.[0-9][0-9][0-9]$//' | uniq > XXX
    for XXX in `cat XXX`; do
        nnn=`ls $XXX.[0-9][0-9][0-9] | tail -1 | sed 's/^.*\./\1/'`
        for i in `seq 0 $((nnn-1))`; do
            rm $XXX.i
        done
    done

2. For each file, check if there is higher number, and delete if yes:

    for i in *.[0-9][0-9][0-9]; do
        XXX=${i%.*}
        nnn=${i#*.}
        if [ -f $XXX.$((nnn+1)) ]; then
            rm $XXX.$nnn
        fi
    done

--

Linux solution for data management and processing.