List only file names in directories and subdirectories in bash

Question

I need to create a list of files in a directory and all of its subdirectories. I've used

find . -type f

to get the list of files, but I only need the file name, not the path that leads to it.

So instead of returning this:

./temp2/temp3/file3.txt
./temp2/file2.txt
./file.txt

I need to return

file3.txt
file2.txt
file.txt

Answer 1

find . -type f -exec basename {} \;

或更好：

find . -type f -printf "%f\n"

Answer 2

You can use printf option in gnu find :

find . -type f -printf "%f\n"

For non-gnu find use -execdir and printf :

find . -type f -execdir printf "%s\n" {} +

Answer 3

find . -type f | xargs basename

The command basename strips the directories and outputs only the file name. Use xargs to chain the output of find as the input of basename .

Answer 4

Late answer :)

The -printf "%f" is the best solution, but the printf is available only in the GNU find. For example on the OS X (and probably FreeBSD too) will print:

find: -printf: unknown primary or operator

for such cases, myself prefer the following

find . -type f | sed 's:.*/::'

It is faster (on the large trees) as the multiple basename execution.

Drawback, it will handle only filenames without the \\n (newline) in the filenames. For handling such cases the easiest (and most universal - but slow) way is using the basename .

You can also use perl, like:

perl -MFile::Find -E 'find(sub{say if -f},".")'