James or anyone, I've been in the habit of using find -ls to record all files in a given hierarchy. Of course, parsing the output of that becomes problematic when a file name contains a newline or other special characters. So what I really want is -ls0, where the entries are delimited with nulls instead of newlines.
In the case of symlinks, I think the " -> " suffices to separate the symlink name from the target name. That is, I'm willing to accept the lossage if a filename contains that string. Less likely than a newline, anyway, and anything involving nulls would greatly complicate the parsing of the output. Oh, maybe it should be \n->\n, that would be even less probable. I thought I would be able to use -printf to get the equivalent, but this appears to be impossible. Aside from formatting trivialities, I don't see a directive or option or other setting to print the exact filename the way that -print0 does. Am I missing something obvious? Is there a way for -printf to print the raw file name? Or the escaped (as in -ls) filename, for that matter? Also, I don't see a way to get the " -> " in symlink entries (without running a separate find for symlinks only), but I guess it's ok for my purposes to just unconditionally use %l. The closest I could get was this: find -printf "%i %k %M %n %u %g %s %T+ %p\n->\n%l\0" BTW, if/when there is an equivalent way to do it using -printf, I think it would be good to state it in the manual (Node: Print File Information), as well as mentioning `ls -dils'. Thanks, Karl
