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Is there a command or flag to clone the user/group ownership and permissions on a file from another file? To make the perms and ownership exactly "like" that of another file?

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3 Answers 3

up vote 31 down vote accepted

On GNU/Linux chown and chmod have a --reference option

chown --reference=otherfile thisfile
chmod --reference=otherfile thisfile
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I never knew this. Sweet, thank you. –  Harv Sep 19 '11 at 17:09
    
this is cool... :) –  greenmang0 Sep 23 '11 at 7:15
    
Could you reference to this answer (and likely cite it) as answer to my question : unix.stackexchange.com/questions/44253/… ? , I think I will be great addition and I'd love to find up-votes there for it. –  Grzegorz Wierzowiecki Jul 31 '12 at 20:40
    
@GrzegorzWierzowiecki: probably that question should be closed, but is a little bit different than this and already has answers, so I better do nothing. –  enzotib Jul 31 '12 at 20:54
    
As you wish and suggest. Thanks for help, I have never put attention to --reference parameter of chmod and chown before :). –  Grzegorz Wierzowiecki Jul 31 '12 at 22:02
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On any unix with GNU utilities, such as (non-embedded) Linux or Cygwin, you can use chmod --reference and chown --reference.

If your system has ACLs, try the ACL commands getfacl and setfacl. These commands differ a little from system to system, but on many you can use getfacl other_file | setfacl -bnM - file_to_change to copy the permissions. This doesn't copy the ownership; you can do that with careful parsing of ls -l other_file, assuming that you don't have user or group names containing whitespace.

LC_ALL=C ls -l other_file | {
  read -r permissions links user group stuff;
  chown -- "$user:$group" file_to_change
}
getfacl other_file | setfacl -bnM - file_to_change
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You should have ACL installed and filesystem mounted with ACL enabled. –  enzotib Sep 14 '11 at 4:03
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@enzotib At least on Linux, ACL tools will work to copy permissions (but not ownership) even if the source and target filesystem don't support ACLs. –  Gilles Sep 14 '11 at 6:57
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If you are not using a system with GNU's chmod/chown (which support the --reference option) you could try to parse the output of ls -l

Here a small script for chmod (if you have a see which supports extended regexes they could be written in a much more readable way ...)

#!/bin/sh

reference=$1
shift
files=$*

# strip the permissions (whith extended regexes could be more readable)
OWNER=$(ls -l ${reference} | sed -e "s/.\(...\).*/\1/"       | sed -e "s/[-]//g" )
GROUP=$(ls -l ${reference} | sed -e "s/....\(...\).*/\1/"    | sed -e "s/[-]//g" )
OTHER=$(ls -l ${reference} | sed -e "s/.......\(...\).*/\1/" | sed -e "s/[-]//g" )

chmod u=${OWNER},g=${GROUP},o=${OTHER} ${files}

UPDATE:

This is even easier using stat:

chmod $( stat -f '%p' ${reference} ) ${files}
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Instead of parsing ls -l output, you could you could parse statoutput. –  jfgagne Sep 21 '11 at 19:51
    
@jfgagne: thanks makes sense I do not know why I didn't think about it in the first place. I updated the answer –  Matteo Sep 22 '11 at 5:28
    
You're using *BSD stat syntax here. Your chmod $(stat ...) command won't work because %p alone outputs too much information for *BSD's chmod, use %Lp to output just the u/g/o bits. Something slightly more elaborate would be required for sticky/setuid/setgid bits. –  mr.spuratic Jun 7 '13 at 10:17
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