Is it possible to define a URI representing a directory path only (i.e. with no actual file terminating the URI)? Example: file:///path/to/files.
I have the need to specify paths to directories to a Java program and would like to specify them as a URI rather than just a String directory path. The program receiving this information would treat the URI as a directory specifier and expect to read whatever files are contained within. It does not care what the file names are.
UPDATE:
If this is allowed, would I be able to "open" the URI in order to get the listing of files within, and then be able to open them for reading?
Yes it is possible.
here's an exemple
lets say myfolder is a folder that contains 3 files (file1.txt, file2.txt, file3.txt).
URI uri = new URI("file:///c:/myfolder/");
File folder =new File(uri);
for(File file : folder.listFiles()){
System.out.println(file.getName());
}
the output will be :
file1.txt
file2.txt
file3.txt
Related
I'm trying to use a file in my code but I don't want to have specify the absolute file path, only the file name, for example "fileName.txt".
I want to do this so I have the ability to use this code on different laptops where the file may be stored in different folders.
The code below is what I'm using at the moment but I receive a NoSuchFileException when I ran it.
FileSystem fs FileSystems.getDefault();
Path fileIn = Paths.get("fileName.txt");
Any ideas how to overcome this problem so I can find the file without knowing its absolute path?
Ideas on how to find the file without knowing its absolute path:
Instruct the user of the app to place the file in the working directory.
Instruct the user of the app to give the path to the file as a program argument, then change the program to use the argument.
Have the program read a configuration file, found using options 1 or 2, then instruct the user of the app to give the path to the file in the configuration file.
Prompt the user for the file name.
(Not recommended) Scan the entire file system for the file, making sure there is only one file with the given name. Optional: If more than one file is found, prompt the user for which file to use.
if you don't ask the user for the complete path, and you don't have a specific folder that it must be in, then your only choice is to search for it.
Start with a rootmost path. Learn to use the File class. Then search all the children. This implementation only returned the first file found with that name.
public File findFile(File folder, String fileName) {
File fullPath = new File(folder,fileName);
if (fullPath.exists()) {
return fullPath;
}
for (File child : folder.listFiles()) {
if (child.isDirectory()) {
File possible = findFile(child,fileName);
if (possible!=null) {
return possible;
}
}
}
return null;
}
Then start this by calling either the root of the file system, or the configured rootmost path that you want to search
File userFile = findFile( new File("/"), fileName );
the best option, however, is to make the user input the entire path. There are nice file system browsing tools for most environments that will do this for the user.
Although the Title isn't very understandable I do have a simple issue. So i'm trying to write some code in a Processing Sketch (https://processing.org/) which can count how many files are in a document. The problem is, is that it doesn't accept the variable type.
File folder = File("My File Path");
folder.listFiles().size;
It says the function File(String) doesn't exist. When I try to put the file path without quation marks, it still doesn't work!
If you have a solution then please use a functioning example so that I know how it works. Thanks for any help!
As Joakim Danielson says it is constructor so you need to use new keyword.
Below code will work for you.
File folder = new File("My File Path");
int fileLength = folder.listFiles().length;
It's a constructor so you need to use new
File folder = new File("My File Path");
//To get the number of files in the folder
folder.listFiles().length;
Assuming the "My File Path" folder is inside your sketch you need to provide the path to your sketch. Luckily Processing already provides a helper function: sketchPath()
Here's an example:
File folder = new File(sketchPath("My File Path"));
println("folder.exists: " + folder.exists());
if(folder.exists()){
println(folder.listFiles().length + " files and/or directories");
}else{
println("folder does not exist, double check the path");
}
Bare in mind there's also a dataPath() function which points to a folder named data in your sketch folder. The data folder is typically used for storing external data (e.g. assets (raster or vector images/Processing font files) or raw data (binary/text/csv/xml/json/etc.)). This is useful to separate your sketch source files from the data to be loaded/accessed by your sketch.
Also, Processing has a few utility functions for listing files and folders.
Be sure to check out Processing > Examples > Topics > File IO > DirectoryList
The example includes less documented functions such as listFiles() (which returns an array of java.io.File objects based on the filters set) or listPaths (which returns an array of String objects: just the paths).
The options and filters are quite handy, for example if you want to list directories only and ignore files you can simply write simply like:
println("directories: " + listFiles(sketchPath("My File Path"),"directories").length);
For example if want to list all the wav files in a data/audio directory inside the sketch you can use:
File[] files = listFiles(dataPath("audio"), "files", "extension=wav");
This will ignore directories and any other file that does not have .wav extension.
To make this answer complete, here are a few more details on the options for listFiles/listPaths from Processing's source code:
"relative" -> no effect with the Files version, but important for listPaths
"recursive"-> traverse nested directories
"extension=js" or "extensions=js|csv|txt" (no dot)
"directories" -> only directories
"files" -> only files
"hidden" -> include hidden files (prefixed with .) disabled by default
Hello one of the parts of a program i'm working on requires the ability to search through a directory. I understand how using a path variable works and how to get to a directory; but once you are in the directory how can you distinguish files from one another? Can we make an array/or a linked list of the files contained within the directory and search using that?
In this specific program the goal is for the user to input a directory, from there go into sub-directory and find a file that ends with .mp3 and copy that to a new user created directory. It is certain that there will only be one .mp3 file in the folder.
Any help would be very much appreciated.
Seeing what you say, I will suppose that you use the java7 Path api.
To know if a path is a directory or a simple file, use Files.isDirectory(Path)
To list the files / directories in your directory, use Files.list(Path)
The javadoc of the Files class : http://docs.oracle.com/javase/8/docs/api/java/nio/file/Files.html
If you use the "old" java.io.File api, then you have a listFiles method, which can take a FileFilter as argument to filter, for exemple, only the files ending with ".mp3".
Good luck
Get Files as so:
List<String> results = new ArrayList<String>();
File[] files = new File("/path/to/the/directory").listFiles();
for (File file : files)
if (file.isFile())
results.add(file.getName());
If you want the extension:
public static String getExtension(String filename){
String extension = "";
int i = fileName.lastIndexOf('.');
if (i > 0)
extension = fileName.substring(i+1);
return extension;
}
There are other ways to get file extensions listed here but they usually require a common external library.
You can use the File object to represent your directory, and then use the listFiles() (which return an array of files File[]) to retrieve the files into that directory.
If you need to search through subdirectories, you can use listFiles() recursively for each directory you encounter.
As for the file extension, the Apache Commons IO library has a neat FileFilter for that: the SuffixFileFilter.
In my app, I used this code:
File DirectoryPath = cw.getDir("custom", Context.MODE_PRIVATE);
While creating a directory, and it returns:
/data/data/com.custom/app_custom**
So my question is why this app_ appears along with directory name. I know its default, but what actually it means?
And secondly, how can I create a sub-directory inside my directory i.e. app_custom in this case. if anyone knows please help me to understand this concept of getDir.
As far as I think, automatic "app_" added to user created data folders avoid any conflicts with system predefined application folders (folders inside application data folder i.e. cache, contents, databases etc. which are automatically created).
One method to create a sub folder inside those "app_..." folders, get absolute path of "app_..." folder, append required folder name to that and create using mkdirs()
e.g.
File dir = new File(newFolderPath);
dir.mkdirs()
Note: sub folders do not get "app_..." prefix
You can create a new Directory using the path that you are getting from getDir(),
File file = getDir("custom", MODE_PRIVATE);
String path = file.getAbsolutePath();
File create_dir = new File(path+"/dir_name");
if(!create_dir.exists()){
create_dir.mkdir();
}
The code I am running is in /Test1/Example. If I need to read a .txt file in /Test1 how do I get Java to go back 1 level in the directory tree, and then read my .txt file
I have searched/googled and have not been able to find a way to read files in a different location.
I am running a java script in an .htm file located at /Test1/Test2/testing.htm. Where it says script src=" ". What would I put in the quotations to have it read from my file located at /Test1/example.txt.
In Java you can use getParentFile() to traverse up the tree. So you started your program in /Test1/Example directory. And you want to write your new file as /Test1/Example.txt
File currentDir = new File(".");
File parentDir = currentDir.getParentFile();
File newFile = new File(parentDir,"Example.txt");;
Obviously there are multiple ways to do this.
You should be able to use the parent directory reference of "../"
You may need to do checks on the OS to determine which directory separation you should be using ['\' compared to '/']
When you create a File object in Java, you can give it a pathname. You can either use an absolute pathname or a relative one. Using absolutes to do what you want would require:
File file = new File("/Test1/myFile.txt");
if(file.canRead())
{
// read file here
}
Using relatives paths if you want to run from the location /Test1/Example:
File file = new File("../myFile.txt");
if(file.canRead())
{
// read file here
}
I had a similar experience.
My requirement is: I have a file named "sample.json" under a directory "input", I have my java file named "JsonRead.java" under a directory "testcase". So, the entire folder structure will be like untitled/splunkAutomation/src and under this I have folders input, testcase.
once after you compile your program, you can see a input file copy named "sample.json" under a folder named "out/production/yourparentfolderabovesrc/input" and class file named "JsonRead.class" under a folder named "out/production/yourparentfolderabovesrc/testcase". So, during run time, Java will actually refer these files and NOT our actual .java file under "src".
So, my JsonRead.java looked like this,
package testcase;
import java.io.*;
import org.json.simple.JSONObject;
public class JsonRead{
public static void main(String[] args){
java.net.URL fileURL=JsonRead.class.getClass().getResource("/input/sample.json");
System.out.println("fileURL: "+fileURL);
File f = new File(fileURL.toURI());
System.out.println("fileIs: "+f);
}
}
This will give you the output like,
fileURL: file:/C:/Users/asanthal/untitled/out/production/splunkAutomation/input/sample.json
fileIs: C:\Users\asanthal\untitled\out\production\splunkAutomation\input\sample.json
It worked for me. I was saving all my classes on a folder but I needed to read an input file from the parent directory of my classes folder. This did the job.
String FileName = "Example.txt";
File parentDir = new File(".."); // New file (parent file ..)
File newFile = new File(parentDir,fileName); //open the file