John W. Krahn wrote:
> Gan Uesli Starling wrote:
>>
>> Am wanting to get a list of files in a directory, but only
>> files which are not themselves further sub-directories.
>>
>> When I do this...
>>
>> opendir(DIR, $this_path) or die "Can't opendir $this_path: $!";
>
> You are opening the directory with opendir() but you are not reading
> the files with readdir().
>
>> @file_list = glob("*");
>
> If you use glob() then you don't have to use opendir() and closedir().
>
>
>> closedir(DIR);
>>
>> ...it gives all the files, including the files which are also
>> directories. Now on UNIX I can get a long list of all files, then
>> grep out the directories, then sed out all but the names on every
>> line. That works like so on NetBSD...
>>
>> $file_list = `ls -l | grep -v "^d" | sed 's/.* //g'`;
>> @file_list = split /\n/, $file_list;
>> shift @file_list;
>>
>> ...which does exactly what I want. It lists the files which are not
>> directories.
>>
>> But I am writing in Perl so as to NOT be OS dependent. I looked into
>> the
>> -d test. But it seems an awful kludge to have to open every file and
>> get a status on it.
>
> You don't have to open the file just to use stat() or the file test
> operators.
>
>
>> Is there a pure Perl way to get a list of file names exclusive of any
>> sub-directory names that will work multi-platform?
>
> Using opendir(), readdir() and closedir():
>
> opendir DIR, $this_path or die "Can't opendir $this_path: $!";
> my @file_list = grep !-d, readdir DIR;
> closedir DIR;
>
Something that always catches me, is that the readdir only pulls the
directory or filename and if you have not done a chdir, then you need to do
some concatenating otherwise your tests will not give you thre results you
want. So you could make change:
@file_list = grep !-d ($this_path . $_), readdir DIR;
where I make the $this_path have the separator / as last character
One way.
Wags ;)
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