filenames

I use Perl on windows(Active Perl). I have a perl program to glob the files in current folder, and concatenate them all using dos copy command called from within using system()... When i execute, this gives a dos error saying "The system cannot find the file specified." It's related to the spaces in the filenames I have. This is the perl code :- @files = glob "*.mp3"; $outfile = 'final.mp3'; $firsttime = 1; foreach (@files) { if($firsttime == 1) { @args = ('copy' ,"/b ","$_","+","$outfile", "$outfile"); system (@args); #system("copy /b '$_'+$outfile $outfile"); $firsttime = 0; } else { @args =

For example, having: <script type="text/javascript" src="http://somedomain.com/js/somejs.js?14"> </script> So what does "?14" means here? Its a url param like any other parameter passed in a url. Sometimes JS scripts are created on the fly using server side technologies other times it is simply a version number to help with browser caching issues. They are there to fool browsers into thinking that it is a new file. This is a trick to avoid browser-cached copy when you update the JS file. The javascript script is probably generated by a server side script (PHP, CGI, etc.) , which takes 14 as a

I am starting with a file containing a list of hundreds of files (full paths) in a random order. I would like to list the details of the ten latest files in that list. This is my naive attempt: $ ls -las -t `cat list-of-files.txt` | head -10 That works, so long as none of the files have spaces in, but fails if they do as those files are split up at the spaces and treated as separate files. File "hello world" gives me: ls: hello: No such file or directory ls: world: No such file or directory I have tried quoting the files in the original list-of-files file, but the here-document still splits

In a C program, I have a file path in a string (specifically, this is the exe name stored in argv[0] ). I would like to extract the file name and discard the directory path using MS Visual Studio 2005. Any built in function for that? Not really, just locate the last backslash in the path. Anything after that is the filename. If nothing is after that, the path specifies a directory name. // Returns filename portion of the given path // Returns empty string if path is directory char *GetFileName(const char *path) { char *filename = strrchr(path, '\\'); if (filename == NULL) filename = path; else

open(my $fh, '>', $path) || die $!; my_sub($fh); Can my_sub() somehow extrapolate $path from $fh? A filehandle might not even be connected to a file but instead to a network socket or a pipe hooked to the standard output of a child process. If you want to associate handles with paths your code opens, use a hash and the fileno operator, e.g. , my %fileno2path; sub myopen { my($path) = @_; open my $fh, "<", $path or die "$0: open: $!"; $fileno2path{fileno $fh} = $path; $fh; } sub myclose { my($fh) = @_; delete $fileno2path{fileno $fh}; close $fh or warn "$0: close: $!"; } sub path { my($fh) = @_

i have a largish C++ project, with source-files organised in multiple folders (on the filesystem). in two of these folders, i have files with the same name. e.g. \MyProject\foo\Blurp.cpp \MyProject\foo\File.cpp \MyProject\bar\File.cpp \MyProject\bar\Knoll.cpp the project is cross platform, and i use autoconf on linux and OSX, but have to use MSVC on W32 (due to some 3rd party C++ libraries i use on W32 and the C++ binary interface incompatibilities across compilers) on the MSVC side, the project is organized into multiple "Filters" (those virtual Folders) as well (with names roughly