Remove path from find command output

Question

I have a bash script for deploying code from a beta environment to a production environment but currently I have to add the list of files to a txt file manaully and sometime I miss some. Basically my deployment script cat/loops copying the files over. (exports/imports db as well but that's not relevant..lol)

Anyway, I'd like to use the find command to generate a list of files modified in the last 14 days. The problem is I need to strip the path out ./ in order for the deployment script to work.

Here's an example of the find command usage:

find . -type f -mtime -14 > deploy.txt

Here's the line that cats deploy.txt in my deployment script:

for i in `cat deploy.txt`; do cp -i /home/user/beta/public_html/$i /home/user/public_html/$i; done

Any idea how to accomplish this using bash scripting?

Thanks!

Is there a reason you can't just use rsync for this? It sounds as if you may be trying to reinvent the wheel and make it a triangle. — ThatGraemeGuy, Jan 27 '12 at 21:05
valid point however before deployment the deploy.txt is reviewed and anything that needs fixing gets removed — Mikey1980, Jan 27 '12 at 21:08
Maybe run rsync in `--dry-run` mode and output that to a file for review? — Zoredache, Jan 27 '12 at 21:14

score 48 · Accepted Answer · edited Feb 23 '18 at 19:00

48

You can use the -printf command line option with %f to print just the filename without any directory information

find . -type f -mtime -14 -printf '%f\n' > deploy.txt

or you can use sed to just remove the ./

find . -type f -mtime -14 | sed 's|^./||' >deploy.txt

edited Feb 23 '18 at 19:00

Vladimir Panteleev

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answered Jan 27 '12 at 21:09

user9517

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I believe you have to escape the `/.`. – EEAA Jan 27 '12 at 21:11
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Not when you use `|` as the delimiter. – user9517 Jan 27 '12 at 21:13
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Using `%f` will strip off all directories, not just `./`. This will break his deployment script if any of the files returned by `find` are located in subdirectories. – James Sneeringer Jan 27 '12 at 21:13
No kidding. That's good to know. :) – EEAA Jan 27 '12 at 21:14
@JamesSneeringer: I know but it's not entirely clear what the OP wants and I did mention that in my answer too – user9517 Jan 27 '12 at 21:15
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@Iain You do need to escape the `.` though. And maybe best to use a start-of-line anchor: `s|^\./||` – James O'Gorman Jan 27 '12 at 21:25
@Iain: So you did, my bad. :) – James Sneeringer Jan 27 '12 at 21:37
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Use `%P` instead of `%f`. This just cut off the path given in the command line. – TrueY May 29 '13 at 10:05
This seems like a nice, but Linux-only trick. Regular `find(1)` -- such as that found on FreeBSD, for example, does not have the `-printf` predicate... – Mikhail T. Mar 22 '18 at 20:26

score 14 · Answer 2 · answered Jan 27 '12 at 21:36

14

The ./ should be harmless. Most programs will treat /foo/bar and /foo/./bar as equivalent. I realize it doesn't look very nice, but based on what you've posted, I see no reason why it would cause your script to fail.

If you really want to strip it off, sed is probably the cleanest way:

find . -type d -mtime 14 | sed -e 's,^\./,,' > deploy.txt

If you're on a system with GNU find (e.g. most Linux systems), you can do it in one shot with find -printf:

find . -type d -mtime 14 -printf "%P\n" > deploy.txt

The %P returns the complete path of each file found, minus the path specified on the command line, up to and including the first slash. This will preserve any subdirectories in your directory structure.

answered Jan 27 '12 at 21:36

James Sneeringer

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The `./` prefix is actually a feature; without it, piping output to other commands will not work correctly e.g. for file names which start with a dash. – tripleee Nov 10 '17 at 05:21
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@tripleee - I understand the utility of it, but OP specifically asked how to strip it off because it was causing problems for his deployment script. Sure, there might be better ways to do what he needs to do, but that's a rabbit hole well beyond the scope of what was asked. – James Sneeringer Nov 18 '17 at 16:32
`rsync` for example won't in some situations when you use `find` to cretae pattern files. – Mecki Sep 28 '20 at 22:03

score 7 · Answer 3 · answered May 29 '13 at 10:01

Why do you need to strip off the ./? It is a valid to have in a path. So

cp -i dir1/./somefile dir2/./somefile

is just ok!

But if You would like to strip off the directory name in find You can use the %P arg to -printf.

man find(1) says:

         %P     File's name with the name of the  command  line  argument
                 under which it was found removed.

An example

$ find other -maxdepth 1
other
other/CVS
other/bin
other/lib
other/doc
other/gdbinit
$ find other -maxdepth 1 -printf "%P\n"

CVS
bin
lib
doc
gdbinit

Mind the first empty line! If You want to avoid it use -mindepth 1

$ find other -mindepth 1 -maxdepth 1 -printf "%P\n"
CVS
bin
lib
doc
gdbinit

Yes, it is valid when specifying paths to be interpreted by the system but it will change behavior when using the find output to generate file list for other tools, e.g. `rsync` include/exclude lists. — Mecki, Sep 28 '20 at 22:02

score 7 · Answer 4 · answered May 26 '14 at 16:12

"find -printf" solution won't work on FreeBSD because find don't have such an option. In this case AWK can help. It returns a last name ($NF), so it can work on any depth.

find /usr/local/etc/rc.d -type f | awk -F/ '{print $NF}'

PS: taken from D.Tansley "Linux and Unix shell programming" book

score 3 · Answer 5 · answered Jan 27 '12 at 21:06

3

Well you have a few options. You could use the -printf option in find to only print out the filename, or you could use a tool like sed to simply strip out the ./.

# simply print out the filename, will break if you have sub-directories.
find . -mtime -14 -printf '%f\n'

# strip a leading ./
find .  -mtime -14 | sed -e 's/^\.\///'

answered Jan 27 '12 at 21:06

Zoredache

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Can you give me an example of it's usage? Sorry, linux is not really my forté – Mikey1980 Jan 27 '12 at 21:11
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Look again, there are examples. – Zoredache Jan 27 '12 at 21:13
Agreed - just try it, you're not going to hurt anything running a combination of find and sed. – EEAA Jan 27 '12 at 21:15

score 3 · Answer 6 · answered Jan 27 '12 at 21:12

A quick solution, if I understand correctly the question, is using cut on the output of the find command:

$> find . -type f -mtime -14 | cut -b 3- > deploy.txt

This will strip the first to characters from each line of the result (in your case ./). Probably not the best solution, but works in your scenario.

score 1 · Answer 7 · answered Jan 27 '12 at 21:08

1

sed is perfect for this sort of thing.

$ find . -type f -mtime -14 find . | sed 's/^\.\///' > deploy.txt

answered Jan 27 '12 at 21:08

EEAA

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Small tip: `sed` allows you to use any other character instead of `/` as separator. When working with paths, this is pretty useful as it requires less escaping: `sed 's|^\./||'` – Mecki Jan 07 '20 at 11:05

score 1 · Answer 8 · answered Feb 21 '17 at 17:59

I don't know what kind of filenames we're talking about, but anything with spaces or newlines puts most of the solutions here at risk.

My suggestion would be to use shell's parameter expansion to strip of the characters from each filename:

find . -mtime -14 -exec sh -c 'printf "${0#./}\n"' {} \; >deploy.txt

If you really really like pipes, you could use a "safe" delimiter, like a null, with xargs to receive each filename:

find . -mtime -14 -print0 | xargs -0 -n 1 sh -c 'printf "${0#./}\n"' >deploy.txt

Remember that the output of this is NOT null-delimited, so while it may be sufficient for an eyeball check, none of these solutions produce a deploy.txt file that is safe to use for automation, unless you're very confident about your source filenames.

Remove path from find command output

8 Answers8