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[[Anchor(faq87)]] == How can I get the permissions of a file without parsing ls -l output? == There are several potential ways, most of which are system-specific. They also depend on precisely ''why'' you want the permissions. The majority of the cases where you might ask this question -- such as ''I want to find any files with the setuid bit set'' -- can be answered by the information in [:UsingFind#permissions:]. As the page name implies, those answers are based on the `find(1)` command. For some questions, such as ''I want to make sure this file has 0644 permissions'', you don't actually need to ''check'' what the permissions are. You can just use `chmod 0644 myfile` and set them directly. If your needs aren't met by any of those, then we can look at a few alternatives: * On GNU/Linux systems, and possibly others, there is a command called `stat(1)`. On older GNU/Linux systems, this command take no options -- just a filename -- and you will have to parse its output. {{{$ stat / File: "/" Size: 1024 Filetype: Directory Mode: (0755/drwxr-xr-x) Uid: ( 0/ root) Gid: ( 0/ root) Device: 8,0 Inode: 2 Links: 25 Access: Wed Oct 17 14:58:02 2007(00000.00:00:01) Modify: Wed Feb 28 15:42:14 2007(00230.22:15:49) Change: Wed Feb 28 15:42:14 2007(00230.22:15:49)}}} In this case, one could extract the 0755 from the `Mode:` line, using `awk` or similar commands. * On newer GNU/Linux systems, the `stat` command takes arguments which allow you to specify which information you want: {{{$ stat -c %a / 755}}} That's obviously a lot easier to parse. * On systems with perl 5, you can use: {{{ perl -e 'printf "%o\n", 07777 & (stat $ARGV[0])[2]' "$filename"}}} This returns the same octal string that the `stat -c %a` example does, but is far more portable. (And slower.) |
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