Read File Names from a Directory

Getting the filenames of all files in a folder

You could do it like that:

File folder = new File("your/path");
File[] listOfFiles = folder.listFiles();

for (int i = 0; i < listOfFiles.length; i++) {
  if (listOfFiles[i].isFile()) {
    System.out.println("File " + listOfFiles[i].getName());
  } else if (listOfFiles[i].isDirectory()) {
    System.out.println("Directory " + listOfFiles[i].getName());
  }
}

Do you want to only get JPEG files or all files?

Directory.GetFiles: how to get only filename, not full path?

You can use System.IO.Path.GetFileName to do this.

E.g.,

string[] files = Directory.GetFiles(dir);
foreach(string file in files)
    Console.WriteLine(Path.GetFileName(file));

While you could use FileInfo, it is much more heavyweight than the approach you are already using (just retrieving file paths). So I would suggest you stick with GetFiles unless you need the additional functionality of the FileInfo class.

How can I get the list of files in a directory using C or C++?

UPDATE 2017:

In C++17 there is now an official way to list files of your file system: std::filesystem. There is an excellent answer from Shreevardhan below with this source code:

#include <string>
#include <iostream>
#include <filesystem>
namespace fs = std::filesystem;

int main()
{
    std::string path = "/path/to/directory";
    for (const auto & entry : fs::directory_iterator(path))
        std::cout << entry.path() << std::endl;
}

Old Answer:

In small and simple tasks I do not use boost, I use dirent.h. It is available as a standard header in UNIX, and also available for Windows via a compatibility layer created by Toni Ronkko.

DIR *dir;
struct dirent *ent;
if ((dir = opendir ("c:\\src\\")) != NULL) {
  /* print all the files and directories within directory */
  while ((ent = readdir (dir)) != NULL) {
    printf ("%s\n", ent->d_name);
  }
  closedir (dir);
} else {
  /* could not open directory */
  perror ("");
  return EXIT_FAILURE;
}

It is just a small header file and does most of the simple stuff you need without using a big template-based approach like boost (no offence, I like boost!).

How to get file names in a directory one-by-one?

Pathlib.iterdir() offers a generator to iterate through directories, which reduces memory consumption:

import sys
import pathlib
import os

path = '/cache/srtm'
pl = pathlib.Path(path).iterdir()
oslb = os.listdir(path)
print(type(pl))
print (type(oslb))

print ('pathlib.iter: %s' % sys.getsizeof(pl))
print ('os.listdir: %s' % sys.getsizeof(oslb))

Prints:

<class 'generator'>
<class 'list'>
pathlib.iter: 88
os.listdir: 124920

Read file names from directory in Bash

To make the smallest change that fixes the problem:

dir="path to the files"
for f in "$dir"/*; do
  cat "$f"
done

To accomplish what you describe as your desired end goal:

shopt -s nullglob
dir="path to the files"
substrings=( R1 R2 )
for substring in "${substrings[@]}"; do
  cat /dev/null "$dir"/*"$substring"* >"${substring}.out"
done

Note that cat can take multiple files in one invocation -- in fact, if you aren't doing that, you usually don't need to use cat at all.

Batch File : Read File Names from a Directory and Store in Array

Simulating Array

String is the only variable type in batch file. However, arrays can be simulated with several variables of identical name except a trailing numerical ID such as:

Array[1]    Array[2]    Array[3]    Array[4]     etc...

We can store each file name into these variables.

---

Retrieving command output

The first step is to put the command output into a variable. We may use the for /f loop.

for /f %%G in ('dir *.txt /b') do set filename=%%~G

The ('') clause and /f option specifies to collect the command output. Note that the filename you get is always the last one displayed because of variable-overwriting. This can be resolved by appending instead, but is beyond the scope of this answer.

---

Giving ID to files

In this case, I will name the array filename with a trailing ID [n], where n is a numerical ID.

setlocal enableDelayedExpansion
set /a ID=1

for /f "delims=" %%G in ('dir *.txt /b') do (
    set filename[!ID!]=%%~G
    set /a ID+=1
)

set filename
endlocal

Two things to note:

• I have added "delims=" into the loop to ensure it works properly with default delimiters.

• I replaced %ID% with !ID! because of delayed expansion. In short, when delayed expansion is disabled, the entire for loop is phrased before runtime, variables with new values are not updated block(()) structures; if enabled, the said variables are updated. !ID! indicates the need for update in each loop.

Batch File; List files in directory, only filenames?

The full command is:

dir /b /a-d

Let me break it up;

Basically the /b is what you look for.

/a-d will exclude the directory names.

---

For more information see dir /? for other arguments that you can use with the dir command.

read file name from directory

. is for current dir
.. is for one directory up

When using readdir you will get those 2 extra.

I prefer using glob(). That function lets you filter for html files only too

<?php
   $files = glob('dir/*html');
   foreach($files as $file) {
       echo "filename:".$file."<br />";
   }
?>

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