How to Get Long Filename from Argv

How to get long filename from ARGV

I don't know if it is possible to change the argument you recieve on a drag and drop, but you could use the Win32 getLongPathName() function, using the Ruby Win32 bindings

--edit--

Including @peter's solution formatted for readability:

require 'find'
require 'fileutils'
require 'Win32API'
def get_long_win32_filename(short_name)
  max_path = 1024
  long_name = " " * max_path
  lfn_size = Win32API.new("kernel32", 
      "GetLongPathName", ['P','P','L'],'L').call(short_name, long_name, max_path)
  return (1..max_path).include?(lfn_size) ? long_name[0..lfn_size-1] : short_name
end 

ARGV.each do|a|
  puts a
  puts get_long_win32_filename(a)
end

Integer and a filename as two command-line arguments in C

• argc tells you the number of arguments.
• argv is an array of char pointers to c-style strings.

So you can simply print all arguments by:

int main(int argc, char *argv[]) 
{
     int i;
     for (i=0; i<argc; ++i)
     {
         printf("%s\n", argv[i]);
     }
}

You can use atoi to convert (the initial portion of) a string to an integer.

So you can do something like:

int main(int argc,char *argv[]) 
{
     char filename[100];
     int size = 0;
     int i;
     if (argc < 3)
     {
         printf("Too few arguments\n");
         return 0;
     }
     if (strlen(argv[1]) >= 100)
     {
         printf("File name too long\n");
         return 0;
     }
     strcpy(filename, argv[1]);
     size = atoi(argv[2]);
     if (size <= 0)
     {
         printf("Invalid size\n");
         return 0;
     }

     ....
     ....

     return 0;
}

Note that it is usually not necessary to copy file name arguments to another variable unless you are going to modify the value in some way.

unable to get filename from argv[1] in C

argv is of type char **, so argv[1] is of type char *. So that's the type of the variable you want to assign this to.

char *name = argv[1];

You can't declare name as char [] and initialize it with a char *. Only a string literal may be used for initialization.

If you want to make a copy of the string rather than have another pointer to it, you can use strdup which allocates memory for the copied string and copies it over.

char *name = strdup(argv[1]);

Don't forget to free it when you're done with it.

Extract the file name from filename with path which comes from argument

You can use this:

std::string filename = string(argv[1]);
int index = filename.find_last_of("/\\");
std::string input_trace_filename = filename.substr(index+1);

How do I define and pass in a file name to fopen() from command line?

That's what the arguments to main are for:

#include <stdio.h>

int main(int argc, char **argv)
{
    char c;
    FILE *fp;

    if (argc >= 2)
         fp = fopen(argv[1], "w");
    else fp = fopen("file.txt", "w");

    while ((c = getchar()) != EOF)
    {
        putc(c, fp);
    }

    return 0;
}

If you followed this, you might wonder what is in argv[0]. That's where the program name is. Some operating system environments put the full path to the executable file there. Others put only the program name. Others still put what was typed.

For the command ../../bin/someprogram

on Windows, argv[0] is "C:\\Documents and Settings\\User\bin\\someprogram.exe"
on Linux/bash, argv[0] is ../../bin/someprogram
on Ultrix/csh, (I think) argv[0] is /home/username/bin/someprogram

How to get file names from command-line parameters in Python

Just use sys.argv:

import sys
import csv

with open(sys.argv[1], "r") as test, open(sys.argv[2], "w") as table:
      # more here

Note, sys.argv[0] contains the script name (in your case, Script.py). To get the first argument, you should get sys.argv[1]; to get the second argument, you should get sys.argv[2] and so on.

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