I need to write a Unix shell script which will list top 10 files in a directory tree on basis of size. i.e. first file should be the biggest in the whole directory and all its sub directories.
Please suggest any ideas

You don't need a "script", it fits into a single line:

find /start/dir -type f -exec du {} \; | sort -rn | head -10

bakunin

Thanks it is listing , But the size that shows when i run ls -l command and the size that is showing for this command is different can you please let me know the reason

example :

$ ll /pfta02/10.001
-rw-rw-rw- 1 pnt pf 87843360 Feb 6 02:21 /pfta02/10.001

$ find /pfta03 -type f -exec du {} \; |sort -rn | head -1
171572 /pfta02/10.001

why the size listing is as 171572 instead of 87843360 ... IS any convertion happening ?..

ls -l will give the output of the total directory links , so dont trust it, du is the exact form to get the info, above commands are perfectly true

I tried the command but I am getting following output for all files in the entire dir tree

find: /dir/filename: Value too large to be stored in data type

Any thoughts on this?

Use "du -h", it will report the size in a human readable form.

However, this will make the sort a bit more complex.

To maintain the sort as it is, use "du -k" which reports in kbytes instead of blocks, meaning the output numbers will be half as big.