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I have to find recursively inside a bash script. Since I can invoke the bash script from anywhere, I am forced to use the absolute path. Sometimes the absolute path can get very long, and the output get's tedious in the following:

#!/usr/bin/env bash
find /long/absolute/path -type f | sort

So I did:

#!/usr/bin/env bash
find /long/absolute/path -type f | xargs -L 1 -I @ echo $(basename @) | sort

But this doesn't gives the basename. Could you please find out where I am going wrong?

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You went wrong in two ways:

  1. You're passing semi-arbitrary filenames from find through two pipes -- to xargs and then on to sort. Use null-terminated -print0 from find, if you're able, or embed the actions inside -exec, as maulinglawns demonstrated.

  2. IF you had filenames that never contained any characters from $IFS, your command would still fail because you've asked xargs to execute echo with a command substitution; that command substitution asks basename to output the base name of @, which probably doesn't exist. Closer would have been:

    find /long/absolute/path -type f | xargs -L 1 -I @ basename @ | sort
    

... as that would have allowed xargs to interpolate the filenames as arguments to basename.

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  • Thanks for the answer. Why do you suggest that probably @ didn't exists? The above script file in the question does print out the filenames although with the absolute path, implying that @ is not blank – Porcupine Sep 17 '18 at 19:15
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    when you used command substitution, it ran before xargs filled in @, so basename was looking for a file named exactly @ – Jeff Schaller Sep 17 '18 at 19:24
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This works for me:

find /some/long/path/ -type f -exec basename {} \;| sort
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  • This works. Is \; necessary? Any idea why mine didn't work? – Porcupine Sep 17 '18 at 18:46
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    Yes, it is necessary. Without it you get find: missing argument to '-exec'. I don't know why your variant didn't work, I rarely use xargs myself. – maulinglawns Sep 17 '18 at 18:48
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    be aware of arbitrary filenames being passed from find through the pipe to sort --- touch $'/some/long/path/foo\nbar' ---> bar on one line, foo on the next. – Jeff Schaller Sep 17 '18 at 18:57

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