CodeGym/Help with Java Tasks/Why Can I Not Even Rename This File??

Level 28

Saint Louis

31.03.2023
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11comments

Why Can I Not Even Rename This File??

Question about the task

Under discussion

One step forward and two steps back this feels like... I can't even get the file renaming task verified. I even just copied and pasted someone else's code for that section who for-sure was passing that and it doesn't change the name of the file. What the heck?? So aggravating. I can't imagine what could be wrong. Okay, for giggles, I just ran it through the validator instead of just running the code. When running the code, it told me, "Cannot rename file (filename)". And yet, the validator accepts it. And... when I print the renamed file's absolute path, it hasn't changed. Uggghhhhh...

1. The main method is passed two arguments as inputs.
The first argument is path, which is the path to the directory; the second is resultFileAbsolutePath, which is the name (full path) of an existing file that will store the result.
2. For each file in the path directory and in all of its subdirectories, do the following:
For each file whose size in bytes is NOT greater than 50, do the following:
2.1. Sort them by file name in ascending order. Don't include the path when sorting.
2.2. Rename resultFileAbsolutePath to "allFilesContent.txt" (use the FileUtils.renameFile method and FileUtils.isExist if necessary).
2.3. Sequentially write the contents of each file from step 2.1 to allFilesContent.txt. After the body of each file, write "\n".
All files have the TXT extension.
Use "/" as the path separator.

Requirements:

The file passed as the second argument to main must be renamed to allFilesContent.txt.
You need to create a stream to write to the renamed file.
The contents of every file that is 50 bytes or smaller must be written to the allFilesContent.txt file.
The output stream to the file must be closed.
Don't use static variables.

package com.codegym.task.task31.task3101; /* Iterating through a file tree 2. For each file in the path directory and in all of its subdirectories, do the following: For each file whose size in bytes is NOT greater than 50, do the following: All files have the TXT extension. Use "/" as the path separator. */ import java.io.*; import java.util.ArrayList; import java.util.Collections; public class Solution { //1. The main method is passed two arguments as inputs. //The first argument is path, which is the path to the directory; //the second is resultFileAbsolutePath, which is the name (full path) of an existing file that will store the result. public static void main(String[] args) { //folder = "c:/path/" (for example) File folder = new File(args[0]); System.out.println(folder); File resultFileAbsolutePath = new File(args[1]); System.out.println(resultFileAbsolutePath.getAbsolutePath()); /*ArrayList<String> filesLessThan50 = new ArrayList<>(); for (File file : folder.listFiles()) { //if (file.getTotalSpace() <= 50) { filesLessThan50.add(file.getName()); // } }*/ /*ArrayList<File> filesLessThan50Files = new ArrayList<>(); for (String fileString : filesLessThan50) { filesLessThan50Files.add(new File(fileString)); } //2.1. Sort them by file name in ascending order. Don't include the path when sorting. //Sorted in ascending order by file name Collections.sort(filesLessThan50);*/ //2.2. Rename resultFileAbsolutePath to "allFilesContent.txt" (use the FileUtils.renameFile method and FileUtils.isExist if necessary). File allFilesContent = new File(resultFileAbsolutePath.getParent() + "\\allFilesContent.txt"); FileUtils.renameFile(resultFileAbsolutePath, allFilesContent); System.out.println(resultFileAbsolutePath.getName() + "dog"); /*try { //2.3. Sequentially write the contents of each file from step 2.1 to allFilesContent.txt. After the body of each file, write "\n". //We need to access each file in our filesLessThan50 ArrayList and write their contents to this file. BufferedWriter bufferedWriter = new BufferedWriter(new FileWriter(resultFileAbsolutePath)); for(File fileMember : filesLessThan50Files){ FileInputStream fileInputStream = new FileInputStream(fileMember); BufferedReader reader = new BufferedReader(new InputStreamReader(fileInputStream)); String line; while((line = reader.readLine()) != null){ bufferedWriter.write(line); } bufferedWriter.write("\n"); fileInputStream.close(); reader.close(); } bufferedWriter.close(); }catch(IOException e){ e.getMessage();*/ } }

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Thomas

Level 41 , Bayreuth, Germany

1 April, 07:51

Maybe a profanity but you can't rename to a some filename if a file with that name already exists.

Mike S

Level 28 , Saint Louis, United States

3 April, 14:26

Okay, so I've sort of got that figured out. On my desktop, it is successfully changing the name of the file to what I ask it to using the FileUtils.rename(). But why in the heck isn't it correctly reading the new filename when I do this?

System.out.println(resultFileAbsolutePath.getAbsolutePath());

Mike S

Level 28 , Saint Louis, United States

3 April, 14:35

I mean, that filename doesn't even exist anymore on my desktop after I've renamed it... (before renaming, it was called "FirstFile". After renaming it becomes "allFilesContent"). But it continues to produce "(parent folder)/FirstFile.txt" with the above code...

Thomas

Level 41 , Bayreuth, Germany

3 April, 15:02

you're passing two arguments

// args[0], the root dir
File resultFileAbsolutePath = new File(args[1]);

now you're supposed to rename that file to allFilesContent.txt and you should use the Util class for that. It is working with File objects -> create a File objects that represents the destination file name.

File destination = new File(resultFileAbsolutePath.getParentFile() + "/allFilesContent.txt");

(the other constructor is not accepted by CG, new File(File, String), so just don't try it). Now check if destination already exists (on the file system), if so delete. Finally rename.

Guadalupe Gagnon

Level 37 , Tampa, United States

31 March, 20:22

1 problem I see is that line 55 creates a writer that writes to the original file name (args[1]) and not the new "allFilesContent" file. That may be your only problem.

Guadalupe Gagnon

Level 37 , Tampa, United States

31 March, 20:26

A tip: lines 29 - 40 have redundant code. The loop at 31 uses Files, gets the String names of files 50 or less, adds them to a String List. The loop at 38 then converts the String names back to Files to add them to a File List. You can skip that conversion and make the first loop add passing Files to a File List. ****** I did see a new problem with lines 43 and 56. Because you sort the String List and not the File List at line 43, line 56 is iterating through an unsorted List. This will fail the output order.

Guadalupe Gagnon

Level 37 , Tampa, United States

31 March, 20:37

I'm not 100% sure how the natural sorting of Files behaves. It may be by the file name or by the full path, or even by file size. Either way you can pass in your own custom comparator to sort it the way that you want. Here:

ArrayList<File> filesLessThan50 = new ArrayList<>();

for (File file : folder.listFiles())
   if (file.getTotalSpace() <= 50)
      filesLessThan50.add(file);

filesLessThan50.sort(new Comparator<File>() {
   @Override
   public int compare(File o1, File o2) {
      return o1.getName().compareTo(o2.getName());
   }
});

Mike S

Level 28 , Saint Louis, United States

3 April, 14:33

Okay, so maybe this is the source of my confusion. Why is it writing to the "old file" in line 55? I thought I renamed it! I feel like there's some nuance in the nature of the rename() that's throwing me off...

Guadalupe Gagnon

Level 37 , Tampa, United States

3 April, 15:29

Ya, ignore my thoughts about that (old file -> new file). I was mistaken in the function of the rename() method.

Mike S

Level 28 , Saint Louis, United States

5 April, 13:28

I can't believe I finally passed this task. Probably a new record for obstacles... wrapped my head around recursion better, 'exploring' within folders of folders, played footsie with renaming files which I'm still confused by the console output not doing what it seems like it should be doing. But hey, 28 points of dark matter. That can go towards my mortgage.

Guadalupe Gagnon

Level 37 , Tampa, United States

5 April, 14:33

I personally do not like recursion. While it is a neat little trick, it almost always results in way more memory use than is necessary. That is because each method call goes on the stack with new memory being used for the methods arguments and local variables. Also, too many methods on the stack will result in a stack overflow exception being thrown. The way that I iterate through the files within a directory, and all sub directories, is by using a queue. A queue first in last out type of list. You get the last element you put it each time using the poll() method, which removes the item as well. The algorithm is simply: 1) put the start directory in the queue that you want to iterate through outer loop** 2) pull next directory out of queue inner loop** 3) iterate through each file in the directory 4) check if the file is a directory 4a) if the file is a directory put it in the queue 4b) if not then check it according to the task directions **end inner loop **end outer loop So, in code:

Queue<File> queue = new LinkedList<>();
queue.offer(folder); // add initial folder

while(!queue.isEmpty())
   for (File file : queue.poll().listFiles()) // pull next item, iterate through files
      if (file.isDirectory())
         queue.offer(file); // if it is directory add it to queue
      else if (file.getTotalSpace() <= 50)
         filesLessThan50.add(file); // otherwise check if it is a file to task conditions