查找不排除文件的统计信息

Question

Within a bash script, I am trying to find files based on their octet using stat in combination with find, however I want to skip some files (file1, file2, etc.). However this doesn't appear to be working. Why is this and how I can fix it? Is this the best way of doing this?

$(stat --format %a 2>&1 $(find /example/dir -type f -not \( -name 'file1' -o \
       -name 'file2' -o -name 'file3' -o -name 'file4' \) -prune) | egrep "777|755"

Answer 1

Original question — 777 permission only

If you are looking for files with 777 permission, use find to do that:

find /example/dir -type f -perm 777

If you don't want to include file1 , file2 , file3 or file4 in the output, use grep too:

find /example/dir -type f -perm 777 | grep -Ev 'file[1234]'

If you want the output from stat for those files, then:

find /example/dir -type f -perm 777 | grep -Ev 'file[1234]' | xargs stat --format %a

or:

stat --format %a $(find /example/dir -type f -perm 777 | grep -Ev 'file[1234]')

This is more likely to run into problems if the list of files is huge. You can reinstate the -prune option on any of the find commands as you require. However, running find example/dir -type f and find example/dir -type f -prune made no difference to the result I saw.

Revised question — 777 and 775 permission

If you're looking for 777 or 775 permission, then you need:

find /example/dir -type f -perm +775

This happens to work because there's only one bit different between the 777 and 775 permissions. A more general and extensible solution would use -or operations:

find /example/dir -type f \( -perm 777 -or -perm 775 \)

With changes in the numbers, this could look for 664 or 646 permission without picking up executable files, which -perm +622 would pick up.

Problems in the question code

As to what is going wrong with the code in the question — I am not completely sure.

$ find example/dir -type f
example/dir/a/filea
example/dir/a/fileb
example/dir/b/filea
example/dir/b/fileb
example/dir/c/filea
example/dir/c/fileb
example/dir/filea
example/dir/fileb
$ find example/dir -type f -not \( -name filea -o -name fileb \)
$ find example/dir -type f -not \( -name filea -or -name fileb \)
$ find example/dir -type f \( -name filea -or -name fileb \)
example/dir/a/filea
example/dir/a/fileb
example/dir/b/filea
example/dir/b/fileb
example/dir/c/filea
example/dir/c/fileb
example/dir/filea
example/dir/fileb
$ find example/dir -type f ! \( -name filea -or -name fileb \)
$ find example/dir -type f \( -not -name filea -and -not -name fileb \)
$ 

The -not or ! operator seems to completely mess things up, which I'd not expect. Superficially, this looks like a bug, but I'd have to have a lot more evidence and have to do a lot of very careful scrutiny of the find specification before I claimed 'bug'.

This testing was done with the find on Mac OS X 10.8.3 (BSD), no GNU find .

(Your use of the term 'octet' in the question is puzzling; it is normally used to indicate a byte in network communications, with the more stringent meaning that it is precisely 8 bits which a byte need not be. The permissions are presented in octal and are based on 16 bits, 2 octets, in the inode.)

Answer 2

Use the -perm option to check permissions, in combination with checking filenames.

find /example/dir -type f -not \( -name 'file1' -o -name 'file2' -o -name 'file3' -o -name 'file4' \) -perm 777

You don't need -prune . This is used to prevent descending until certain subdirectories, it doesn't do anything with files. And it applies to directories that match the specification, so using it with -not in your case would be the opposite of what you wanted.