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Re: bash, grep, and regular expressions

Kevin Mark wrote:

On Thu, Feb 17, 2005 at 08:49:41PM -0800, Freddy Freeloader wrote:


Glenn English wrote:


On Thu, 2005-02-17 at 16:08 -0800, Freddy Freeloader wrote:
What I've been attempting to do with grep and regular expressions is list only non-hidden directories and/or files. I am unable to come up with an expression that will elimate hidden files and return non-hidden files at the same time. 

ls -al | grep -v ' \.\<[a-zA-Z0-9].*\>' # returns everything

ls | grep -e '\<[^.][[:alnum:]]'  # returns everything

ls | grep -e '\<[.][[:alnum:]]'  # returns an empty set




ls -al | grep -v ' \.'   seems to work here???
I thought about this a little more and from it's behavior of filtering out files with extensions I'd say it's not filtering based on the . that designates whether a file or directory is hidden.
I have a few files without extensions and it returns those, but any file with an extension is filtered out. So, this isn't really resovling the problem I'm having. 




Hi Freddy,
find -maxdepth 1 |grep -v "^\./\." (remove files that start with ./.) 
this find files and directories in the current directory that are not
hidden. Hidden files and directories start with .
Cheers,
Kev


Hi Kevin,
This returns the correct files ok. However, it's not too much different than ls -a | grep -v '^\.' which I can make work too. What I'm really trying to do is figure out how to parse the beginning of the file name column in the long format, and I just can't come up with a way to do that on my machines. I've been given expressions that do it on other machines, but I have yet to find one that will work on my machines.
I just can't see why grep can't recognize the beginning of the word in the file name column in the long format. Why will it recognize the . in short format at the beginning of a line, but not in the long format at the beginning of a word? I can parse the beginning of words in any other situation I've come across but not in in ls -al.
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