使用regexp从Unix“ls -la”命令获取文件名？

Question

How can I produce a regular expressions pattern that returns the filename from any one of these lines? (I will search one line at a time).

drwxrwxrwx  4 apache      apache       4096 Oct 14 09:40 .
drwxrwxrwx 11 apache      apache       4096 Oct 13 11:33 ..
-rwxrwxrwx  1 apache      apache      16507 Oct 17 10:16 .bash_history
-rwxrwxrwx  1 apache      apache         33 Sep  1 09:36 .bash_logout
-rwxrwxrwx  1 apache      apache        176 Sep  1 09:36 .bash_profile
-rwxrwxrwx  1 apache      apache        124 Sep  1 09:36 .bashrc
-rwxrwxrwx  1 apache      apache        515 Sep  1 09:36 .emacs
-rw-------  1 christoffer christoffer 11993 Sep 18 10:00 .mysql_history
drwxrwxrwx  3 apache      apache       4096 Sep  1 09:48 .subversion
-rwxrwxrwx  1 christoffer christoffer  9204 Oct 14 09:40 .viminfo
drwxrwxrwx 14 apache      apache       4096 Oct 12 07:39 www

The search is done using PHP, but I guess that doesn't really make a difference. :)

EDIT: The file listing is retrieved by a SSH connection and that is why I don't use a built in PHP-function. I need this full listing to see whether or not a file is actually a directory.

Answer 1

Try ls -a1F instead. That will list you all entries (-a), one per line (-1), with additional information about the file type appended to the name (-F).

You will then probably get something like this for your directory:

./
../
.bash_history
.bash_logout
.bash_profile
.bashrc
.emacs
.mysql_history
.subversion/
.viminfo
www/

The directories have a slash / at the end.