listing duplicate files

$ find . -name '*.jpg'
./iPhoto Library/2002/09/30/823167-R1-00-25.jpg
./iPhoto Library/2002/09/30/823167-R1-01-24.jpg
./iPhoto Library/2002/09/30/823167-R1-02-23.jpg
./iPhoto Library/2002/09/30/823167-R1-03-22.jpg
./iPhoto Library/2002/09/30/823167-R1-04-21.jpg
./iPhoto Library/2002/09/30/823167-R1-05-20.jpg
./iPhoto Library/2002/09/30/823167-R1-06-19.jpg
./iPhoto Library/2002/09/30/823167-R1-07-18.jpg
...

I would prefer to see:

823167-R1-00-25.jpg
823167-R1-01-24.jpg
823167-R1-02-23.jpg
823167-R1-03-22.jpg
823167-R1-04-21.jpg
823167-R1-05-20.jpg
823167-R1-06-19.jpg
823167-R1-07-18.jpg
...

Can't use 'cut', don't know how many fields to cut from the output. Some
may have 5 '/' as above, some may have more or less.
Haven't made any progress using 'sed', might work...?

Any ideas?

Thanks,
Charles

Use awk:

awk -F '/' '{ print $NF }'

To count the instances of each:

awk -F '/' '{ ++f[$NF] }
END { for (file in f) printf "%5d %s\n", f[file], f }'

--
Chris F.A. Johnson, author | <http://cfaj.freeshell.org>
Shell Scripting Recipes: | My code in this post, if any,
A Problem-Solution Approach | is released under the
2005, Apress | GNU General Public Licence

find . -name '*.jpg' |

List duplicate files:

find . -name '*.jpg' |
awk -F/ 's[$NF]++==1{print $NF}'

With a counter:

find . -name '*.jpg' |
awk -F/ '{s[$NF]++}END{for(i in s)if(s[i]>1)print s[i],i}'

--
Michael Tosch @ hp : com

$ find . -name '*.jpg' -print -exec basename {} \;

XC

a very slow solution...

for i in `find . -name '*.jpg' -exec basename {} \; ` ; do for j in
`find . -name "$i" -print | wc -l`; do if test "$j" -gt 1; then printf
"%s : %s" "$j" "$i" ; fi; done done

Some more ideas?
It should be nice if we could see filenames of duplicates (directories...)

xavier

find . -name '*.jpg' -printf "%f\n"

Frank

To find duplicate file names, why not try:
find . -name '*.jpg' -exec basename {} \; | sort | uniq -d
or find the number of occurrences for each fine:
find . -name '*.jpg' -exec basename {} \; | sort | uniq -c

XC

How about using fdupes?

--
Saludos,
Ángel

a) this will list all the files in all subdirectories:

find . -name '*.c' -exec basename {} | sort|uniq -c|sort -nr|grep -v '
1'|while read n f; do find . -name $f; done

b) this will show how manu duplicates along with the filename itself:
find . -name '*.c' -exec basename {} | sort|uniq -c|sort -nr|grep -v '
1'

In bash, missing arg. to exec (or am I doing something wrong?)

--
Etienne Marais
Cosmic Link
South Africa

-exec basename {} should be -exec basename {} +

(?)

You need an (escaped) semi-colon:
find . -name '*.c' -exec basename {} \; | whatever

--
John.

No, it shouldn't; basename takes at most two arguments. It should
be: -exec basename {} \;

Besides the missing \; after the -exec,
aren't the following ones faster *and* run on old systems?

b)
find . -name '*.c' -print |
awk -F/ '{s[$NF]++} END {if(NR)for(i in s)if(s[i]>1)print s[i],i}'

a)
find . -name '*.c' -print > tmpf
awk -F/ '{s[$NF]++} END {if(NR)for(i in s)if(s[i]>1)print i}' tmpf |
while read f; do awk -F/ '$NF=="'$f'"' tmpf; done

Correction: you never know what find finds, so it should better be

while read f; do awk -F/ '$NF==F' F="$f" tmpf; done