How to get the filenames and their absolute paths in one line?

Question

I want to list all the files in a folder and their respective absolute paths line by line as below,

filename1  /home/.../file1  
filename2 /home/.../file2  
filename3 /home/.../file3  

I tried with find $(pwd) -type f, but it only gives the paths.

Answer 1

Using GNU find:

find "$PWD" -type f -printf '%f %h/%f\n'

• %f File's name with any leading directories removed (only the last element).
• %h Leading directories of file's name (all but the last element). If the file name contains no slashes (since it is in the current directory) the %h specifier expands to `.'.

Answer 2

With standard find:

find "$PWD" -type f -exec sh -c '
    for pathname do
        printf "%s\t%s\n" "${pathname##*/}" "$pathname"
    done' sh {} +

This would print a tab-delimited list of all files in or below the current directory. The first column would contain the filename of the file, and the second column would contain the pathname of the file relative to the absolute path of the current directory (the $PWD in the search path).

This is done by feeding batches of found pathnames to a short in-line script which simply outputs the names in the preferred format. The parameter substitution ${pathname##*/} expands to the value of $pathname with everything up to the last / removed.

Note that filenames containing newlines would be presented on several lines, and that it would be difficult to distinguish filenames containing tab-characters properly. Assuming that the result would be for viewing rather than parsing, that might be okay.

Answer 3

Hi and thanks for mentioning fselect!

The problem with odd filenames is not new obviously. We can't even be sure that the filename and its path will be on the same line in case they contain \n within. To address those issues fselect supports the INTO option.

The simplest variant is to use INTO LIST for further parsing with xargs. One can be interested in INTO JSON for more elaborate processing.

Answer 4

I would like to mention fselect which is written in Rust and uses a SQL(ish) syntax which I can remember a bit more easily:

fselect name, abspath from .

For illustrating the relevant issue with spaces and new lines as in filename posted by @ilkkachu in comments, here's an example with 2 files: a b c and x\ny\nz:

$ fselect name, abspath from . order by name
a b  c  /path/to/a b  c
x
y
z       /path/to/x
y
z

In fselect there is the json output which might help:

$ fselect name, abspath from . order by name into json
[{"Name":"a b  c","AbsPath":"/path/to/a b  c"},{"AbsPath":"/path/to/x\ny\nz","Name":"x\ny\nz"}]

This find command should work also:

find . -type f -exec basename {} \; -exec realpath {} \; | xargs -n 2

xargs -n2 helps putting the output on the same line (please see comments)

Answer 5

With zsh:

for f (**/*(DN)) print -r -- $f $f:P

$f:P does the equivalent of realpath() so expands to the canonical absolute path of the file (expanding all symlinks in the path components to their target, recursively).

Answer 6

The following bash oneliner will produce the output you want

for f in $(find $(pwd) -type f); do echo "$(basename $f) $(realpath $(dirname $f))"; done

Explanation:
for _ in _ ; do _ ; done iterates over what your command found
basename strips all but the last part from a path (aka. the filename)
dirname does the opposite
realpath returns the actual path (so you don't get . for file in cwd)

If you use this frequently, I suggest aliasing the oneliner to a shorter command.

Answer 7

Use this cmd:

find $PWD -type f -printf '%f %h/%f/n'

or

Looking for some files then go with this:

find $PWD -type f -iname "filenames" -printf '%f %h/%f/n'

Answer 8

You can use the following cmd for searching the files with their absolute paths. E.g. I am trying to search .sh files using cmd as follows:

find . -type f -iname "*.sh"