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getResourceAsStream returning Null when reading file [duplicate]

I am trying to read a text file which is set in CLASSPATH system variable. Not a user variable.
I am trying to get input stream to the file as below:
Place the directory of file (D:\myDir) in CLASSPATH and try below:
InputStream in = this.getClass().getClassLoader().getResourceAsStream("SomeTextFile.txt");
InputStream in = this.getClass().getClassLoader().getResourceAsStream("/SomeTextFile.txt");
InputStream in = this.getClass().getClassLoader().getResourceAsStream("//SomeTextFile.txt");
Place full path of file (D:\myDir\SomeTextFile.txt) in CLASSPATH and try the same above 3 lines of code.
But unfortunately NONE of them are working and I am always getting null into my InputStream in.

With the directory on the classpath, from a class loaded by the same classloader, you should be able to use either of:
// From ClassLoader, all paths are "absolute" already - there's no context
// from which they could be relative. Therefore you don't need a leading slash.
InputStream in = this.getClass().getClassLoader()
.getResourceAsStream("SomeTextFile.txt");
// From Class, the path is relative to the package of the class unless
// you include a leading slash, so if you don't want to use the current
// package, include a slash like this:
InputStream in = this.getClass().getResourceAsStream("/SomeTextFile.txt");
If those aren't working, that suggests something else is wrong.
So for example, take this code:
package dummy;
import java.io.*;
public class Test
{
public static void main(String[] args)
{
InputStream stream = Test.class.getResourceAsStream("/SomeTextFile.txt");
System.out.println(stream != null);
stream = Test.class.getClassLoader().getResourceAsStream("SomeTextFile.txt");
System.out.println(stream != null);
}
}
And this directory structure:
code
dummy
Test.class
txt
SomeTextFile.txt
And then (using the Unix path separator as I'm on a Linux box):
java -classpath code:txt dummy.Test
Results:
true
true

When using the Spring Framework (either as a collection of utilities or container - you do not need to use the latter functionality) you can easily use the Resource abstraction.
Resource resource = new ClassPathResource("com/example/Foo.class");
Through the Resource interface you can access the resource as InputStream, URL, URI or File. Changing the resource type to e.g. a file system resource is a simple matter of changing the instance.

This is how I read all lines of a text file on my classpath, using Java 7 NIO:
...
import java.nio.charset.Charset;
import java.nio.file.Files;
import java.nio.file.Paths;
...
Files.readAllLines(
Paths.get(this.getClass().getResource("res.txt").toURI()), Charset.defaultCharset());
NB this is an example of how it can be done. You'll have to make improvements as necessary. This example will only work if the file is actually present on your classpath, otherwise a NullPointerException will be thrown when getResource() returns null and .toURI() is invoked on it.
Also, since Java 7, one convenient way of specifying character sets is to use the constants defined in java.nio.charset.StandardCharsets
(these are, according to their javadocs, "guaranteed to be available on every implementation of the Java platform.").
Hence, if you know the encoding of the file to be UTF-8, then specify explicitly the charset StandardCharsets.UTF_8

Please try
InputStream in = this.getClass().getResourceAsStream("/SomeTextFile.txt");
Your tries didn't work because only the class loader for your classes is able to load from the classpath. You used the class loader for the java system itself.

To actually read the contents of the file, I like using Commons IO + Spring Core. Assuming Java 8:
try (InputStream stream = new ClassPathResource("package/resource").getInputStream()) {
IOUtils.toString(stream);
}
Alternatively:
InputStream stream = null;
try {
stream = new ClassPathResource("/log4j.xml").getInputStream();
IOUtils.toString(stream);
} finally {
IOUtils.closeQuietly(stream);
}

To get the class absolute path try this:
String url = this.getClass().getResource("").getPath();

Somehow the best answer doesn't work for me. I need to use a slightly different code instead.
ClassLoader loader = Thread.currentThread().getContextClassLoader();
InputStream is = loader.getResourceAsStream("SomeTextFile.txt");
I hope this help those who encounters the same issue.

If you use Guava:
import com.google.common.io.Resources;
we can get URL from CLASSPATH:
URL resource = Resources.getResource("test.txt");
String file = resource.getFile(); // get file path
or InputStream:
InputStream is = Resources.getResource("test.txt").openStream();
Ways to convert an InputStream to a String

To read the contents of a file into a String from the classpath, you can use this:
private String resourceToString(String filePath) throws IOException, URISyntaxException
{
try (InputStream inputStream = this.getClass().getClassLoader().getResourceAsStream(filePath))
{
return IOUtils.toString(inputStream);
}
}
Note:
IOUtils is part of Commons IO.
Call it like this:
String fileContents = resourceToString("ImOnTheClasspath.txt");

You say "I am trying to read a text file which is set in CLASSPATH system variable." My guess this is on Windows and you are using this ugly dialog to edit the "System Variables".
Now you run your Java program in the console. And that doesn't work: The console gets a copy of the values of the system variables once when it is started. This means any change in the dialog afterwards doesn't have any effect.
There are these solutions:
Start a new console after every change
Use set CLASSPATH=... in the console to set the copy of the variable in the console and when your code works, paste the last value into the variable dialog.
Put the call to Java into .BAT file and double click it. This will create a new console every time (thus copying the current value of the system variable).
BEWARE: If you also have a User variable CLASSPATH then it will shadow your system variable. That is why it is usually better to put the call to your Java program into a .BAT file and set the classpath in there (using set CLASSPATH=) rather than relying on a global system or user variable.
This also makes sure that you can have more than one Java program working on your computer because they are bound to have different classpaths.

My answer is not exactly what is asked in the question. Rather I am giving a solution exactly how easily we can read a file into out java application from our project class path.
For example suppose a config file name example.xml is located in a path like below:-
com.myproject.config.dev
and our java executable class file is in the below path:-
com.myproject.server.main
now just check in both the above path which is the nearest common directory/folder from where you can access both dev and main directory/folder (com.myproject.server.main - where our application’s java executable class is existed) – We can see that it is myproject folder/directory which is the nearest common directory/folder from where we can access our example.xml file. Therefore from a java executable class resides in folder/directory main we have to go back two steps like ../../ to access myproject. Now following this, see how we can read the file:-
package com.myproject.server.main;
class Example {
File xmlFile;
public Example(){
String filePath = this.getClass().getResource("../../config/dev/example.xml").getPath();
this.xmlFile = new File(filePath);
}
public File getXMLFile() {
return this.xmlFile;
}
public static void main(String args[]){
Example ex = new Example();
File xmlFile = ex.getXMLFile();
}
}

If you compile your project in jar file:
you can put your file in resources/files/your_file.text or pdf;
and use this code:
import org.slf4j.Logger;
import org.slf4j.LoggerFactory;
import java.io.*;
public class readFileService(){
private static final Logger LOGGER = LoggerFactory.getLogger(readFileService.class);
public byte[] getFile(){
String filePath="/files/your_file";
InputStream inputStreamFile;
byte[] bytes;
try{
inputStreamFile = this.getClass().getResourceAsStream(filePath);
bytes = new byte[inputStreamFile.available()];
inputStreamFile.read(bytes);
} catch(NullPointerException | IOException e) {
LOGGER.error("Erreur read file "+filePath+" error message :" +e.getMessage());
return null;
}
return bytes;
}
}

I am using webshpere application server and my Web Module is build on Spring MVC. The Test.properties were located in the resources folder, i tried to load this files using the following:
this.getClass().getClassLoader().getResourceAsStream("Test.properties");
this.getClass().getResourceAsStream("/Test.properties");
None of the above code loaded the file.
But with the help of below code the property file was loaded successfully:
Thread.currentThread().getContextClassLoader().getResourceAsStream("Test.properties");
Thanks to the user "user1695166".

Use org.apache.commons.io.FileUtils.readFileToString(new File("src/test/resources/sample-data/fileName.txt"));

Don't use getClassLoader() method and use the "/" before the file name. "/" is very important
this.getClass().getResourceAsStream("/SomeTextFile.txt");

import java.io.BufferedReader;
import java.io.File;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
public class ReadFile
{
/**
* * feel free to make any modification I have have been here so I feel you
* * * #param args * #throws InterruptedException
*/
public static void main(String[] args) throws InterruptedException {
// thread pool of 10
File dir = new File(".");
// read file from same directory as source //
if (dir.isDirectory()) {
File[] files = dir.listFiles();
for (File file : files) {
// if you wanna read file name with txt files
if (file.getName().contains("txt")) {
System.out.println(file.getName());
}
// if you want to open text file and read each line then
if (file.getName().contains("txt")) {
try {
// FileReader reads text files in the default encoding.
FileReader fileReader = new FileReader(
file.getAbsolutePath());
// Always wrap FileReader in BufferedReader.
BufferedReader bufferedReader = new BufferedReader(
fileReader);
String line;
// get file details and get info you need.
while ((line = bufferedReader.readLine()) != null) {
System.out.println(line);
// here you can say...
// System.out.println(line.substring(0, 10)); this
// prints from 0 to 10 indext
}
} catch (FileNotFoundException ex) {
System.out.println("Unable to open file '"
+ file.getName() + "'");
} catch (IOException ex) {
System.out.println("Error reading file '"
+ file.getName() + "'");
// Or we could just do this:
ex.printStackTrace();
}
}
}
}
}
}

you have to put your 'system variable' on the java classpath.

How to get File name from cmd argument to create a copy?

I'm working on a program that reads from a file with a custom extension I made. The idea is that an error report is created every time a file is read. The error report must be in whatever folder the source file was called from. The error file is a copy of the source file, but it has a line number at the beginning of each line and indicates at the end of the line if an error occurred at that line.
(I'm not trying to set up the numbering on this question, this question is just about creating the copy)
So for example, when I call my program from the command prompt:
C:\MyLocation>java =jar myJavaProgram.jar myFileToRead.CustomExtension
Asides from reading the file, it should also create a copy at the same location called myFileToRead-ErrorReport.txt
Additionally: If the source file has no extension, I have to assume that it's still the correct extension, so there won't always be a '.myCustomExtension' segment to replace into .txt
The problem is that I don't know how to grab the file name, because it's coming from the args list of the main method. I am using the following to read the file
public static void main(String[] args) throws FileNotFoundException {
try{
File inputFile = new File(args[0]);
Scanner sc = new Scanner(inputFile);
while(sc.hasNext()){
System.out.println(sc.nextLine());
}
} catch (FileNotFoundException ex) {
System.out.println(ex.getMessage());
System.out.println("File not found.");
}
}
So how can I get that file name to make something like
File errorReport = new File("./" + inputFileName + ".txt"); ?

First the code. The explanations appear after the code.
public static void main(String[] args) {
if (args.length < 1) {
System.out.println("ERROR: Missing filename argument.");
}
else {
String filename = args[0];
if (filename.length() == 0) {
System.out.println("ERROR: Empty filename argument.");
}
else {
if (!filename.endsWith(".CustomExtension")) {
filename += ".CustomExtension";
}
String name = filename.substring(0, filename.indexOf(".CustomExtension"));
name += "-ErrorReport.txt";
File inputFile = new File(filename);
File directory = inputFile.getParentFile();
File errorReport = new File(directory, name);
System.out.println(errorReport.getAbsolutePath());
}
}
}
I make it a habit of checking the parameters. Hence I first check that the file name was supplied. If it was, then I check that it is not an empty string. Note that I have omitted some checks, for example checking whether the named file exists and is readable.
You wrote in your question that the file name argument may or may not include the .CustomExtension. Hence I check whether the supplied name ends with the required extension and append it if necessary. Now, since I know what the file name ends with, that means that the required part of the name is everything up to the extension and that's what the call to substring() gives me.
Once I have the required name, I just append the part that you want to append, i.e. -ErrorReport.txt.
Method getParentFile() in class java.io.File returns the directory that the file is located in. Hence I have the directory that the input file is in. Finally I can create the error report file in the same directory as the input file and with the desired file name. For that I use the constructor of class java.io.File that takes two parameters. Read the javadoc for details.
Note that creating a File object does not create the file. Creating an object to write to the file does, for example FileWriter or OutputStreamWriter.

Here is the code example to create a file, with filename passed from cmd line as argument and to get the same file name :
Class Demo{
public static void main(String[]args){
String path ="<path of file>"
String name= args[0];
File f = new File(path+name+".txt");
f.createNewFile(); //create file
System.out.println(f.getName()); // will give you the file name
}
}
cmd line : java -cp . Demo.java <filename>
Note : '.' used in the cmd if your class file is present in current dir
You can refer the code and modify to suit your requirement.
Hope this is what you are looking for.

Error: [filename].txt (No such file or directory) when trying to read in a txt file in Java (Ubuntu cmd line)

So I'm writing a piece of code that reads in many addresses and then manipulates the data, line by line, then adds it to a Binary Search Tree. However, the file isn't being found no matter what I try. I know the txt file needs to be in the root directory, but (I'm a student) we're required to submit a .tar.gz containing the contents of a directory made specifically for the assignment so I suppose it needs to be in the submitted folder.
I've tried using the "normal" methods I used in IDEs, but they all return the same error.
My code (roughly - obviously you don't really need to see the BST/manipulation stuff), right now, but I've tried every other typical file read-in method:
import java.io.InputStream;
import java.io.*;
import java.util.Scanner;
public class PrintIt{
public static void main(String args[]){
try{
File inFile = new File("/bin/testdata.txt");
Scanner sc = new Scanner(inFile);
while (sc.hasNextLine()) {
String line = sc.nextLine();
System.out.println(line); //manipulation in place of this
}
sc.close();
}catch (Exception e){
System.err.println("Error: " + e.getMessage());
}
}
}
I really only need the reading in the file part - the BST stuff/manipulation is just for context, just in case it mattered/there's a better way than reading through each line of the file and adding each to the Tree.

Android Can't Remember My Files

I am writing a notepad app and I'm making a create a password screen pop up until you've created one, and then a log in screen will pop up from then on.
Here is some sample code:
File myFile = new File(getFilesDir() + "pass.txt");
if (!myFile.exists()) // if "pass.txt" DOESN'T exist, make them create a password
{
try {
// this writes the password to the "pass.txt" file
// which is the one that is checked to exist.
// after it is written to, it should always exist.
FileOutputStream fos = openFileOutput("pass.txt", Context.MODE_PRIVATE);
fos.write(pass.getBytes());
// this writes the security question to a different file.
fos = openFileOutput("securityQ.txt", Context.MODE_PRIVATE);
fos.write(secQ.getBytes());
// this writes the security answer to a different file.
fos = openFileOutput("securityAnswer.txt", Context.MODE_PRIVATE);
fos.write(secAns.getBytes());
fos.close();
} catch(Exception e) {}
^ That is in one method. Then, in another I do this:
try { // input the right password to the String data
char[] inputBuffer = new char[1024];
fIn = openFileInput("pass.txt");
isr = new InputStreamReader(fIn);
isr.read(inputBuffer);
data = new String(inputBuffer);
isr.close();
fIn.close();
}catch(IOException e){}
if (password.getText().toString().equals(data)) // if password is right, log in.
{
loggedin();
}
else // if the password entered is wrong, display the right one.
{
TextView scr = (TextView)findViewById(R.id.display);
scr.setText("." + data + "."+'\n'+"." + password.getText().toString() + ".");
}
The problem is that the user can't log in even though the password is entered correctly and the display proves that.
The other problem is that whenever I run the app again, it goes to the create screen, which means that it is recognizes the file to NOT exist (even though I just wrote to it).
I've dealt with files this whole project and it can keep track of entered text so that when you press a button, it reads the file back to you. Even if you close it, it keeps track of what you input. For some reason though, the password thing doesn't work.
Here is an image of what happens (the first .k. is the data read from the "pass.txt" file and the second .k. is the user inputted String from the EditText):
SOLUTION to the logging in problem:
String values look the same but don't ".equals()" each other
Had to simply use the .trim() method on the password user input.

I'll pass on commenting about saving passwords in a file called "pass.txt", and just focus on the technical part.
File myFile = new File(getFilesDir() + "pass.txt");
myFile will never be a valid file. You don't have a separator / between the path and the file name. Since that will never be vaild, the next line will say it doesn't exist and go through the whole block.
You can easily fix this one of two ways:
File myFile = new File(getFilesDir() + "/pass.txt");
That simply adds the separator to the file name.
File myFile = new File(getFilesDir(), "pass.txt");
This is probably the better option, since it uses the explicit path, file constructor. Either one is fine, though.

You can also just use context.openFileInput("pass.txt"); and catch if FileNotFoundException occurs, at which point you can "assume" the file actually doesn't exist.

How to read text file from classpath in Java?

I am trying to read a text file which is set in CLASSPATH system variable. Not a user variable.
I am trying to get input stream to the file as below:
Place the directory of file (D:\myDir) in CLASSPATH and try below:
InputStream in = this.getClass().getClassLoader().getResourceAsStream("SomeTextFile.txt");
InputStream in = this.getClass().getClassLoader().getResourceAsStream("/SomeTextFile.txt");
InputStream in = this.getClass().getClassLoader().getResourceAsStream("//SomeTextFile.txt");
Place full path of file (D:\myDir\SomeTextFile.txt) in CLASSPATH and try the same above 3 lines of code.
But unfortunately NONE of them are working and I am always getting null into my InputStream in.

With the directory on the classpath, from a class loaded by the same classloader, you should be able to use either of:
// From ClassLoader, all paths are "absolute" already - there's no context
// from which they could be relative. Therefore you don't need a leading slash.
InputStream in = this.getClass().getClassLoader()
.getResourceAsStream("SomeTextFile.txt");
// From Class, the path is relative to the package of the class unless
// you include a leading slash, so if you don't want to use the current
// package, include a slash like this:
InputStream in = this.getClass().getResourceAsStream("/SomeTextFile.txt");
If those aren't working, that suggests something else is wrong.
So for example, take this code:
package dummy;
import java.io.*;
public class Test
{
public static void main(String[] args)
{
InputStream stream = Test.class.getResourceAsStream("/SomeTextFile.txt");
System.out.println(stream != null);
stream = Test.class.getClassLoader().getResourceAsStream("SomeTextFile.txt");
System.out.println(stream != null);
}
}
And this directory structure:
code
dummy
Test.class
txt
SomeTextFile.txt
And then (using the Unix path separator as I'm on a Linux box):
java -classpath code:txt dummy.Test
Results:
true
true

When using the Spring Framework (either as a collection of utilities or container - you do not need to use the latter functionality) you can easily use the Resource abstraction.
Resource resource = new ClassPathResource("com/example/Foo.class");
Through the Resource interface you can access the resource as InputStream, URL, URI or File. Changing the resource type to e.g. a file system resource is a simple matter of changing the instance.

This is how I read all lines of a text file on my classpath, using Java 7 NIO:
...
import java.nio.charset.Charset;
import java.nio.file.Files;
import java.nio.file.Paths;
...
Files.readAllLines(
Paths.get(this.getClass().getResource("res.txt").toURI()), Charset.defaultCharset());
NB this is an example of how it can be done. You'll have to make improvements as necessary. This example will only work if the file is actually present on your classpath, otherwise a NullPointerException will be thrown when getResource() returns null and .toURI() is invoked on it.
Also, since Java 7, one convenient way of specifying character sets is to use the constants defined in java.nio.charset.StandardCharsets
(these are, according to their javadocs, "guaranteed to be available on every implementation of the Java platform.").
Hence, if you know the encoding of the file to be UTF-8, then specify explicitly the charset StandardCharsets.UTF_8

Please try
InputStream in = this.getClass().getResourceAsStream("/SomeTextFile.txt");
Your tries didn't work because only the class loader for your classes is able to load from the classpath. You used the class loader for the java system itself.

To actually read the contents of the file, I like using Commons IO + Spring Core. Assuming Java 8:
try (InputStream stream = new ClassPathResource("package/resource").getInputStream()) {
IOUtils.toString(stream);
}
Alternatively:
InputStream stream = null;
try {
stream = new ClassPathResource("/log4j.xml").getInputStream();
IOUtils.toString(stream);
} finally {
IOUtils.closeQuietly(stream);
}

To get the class absolute path try this:
String url = this.getClass().getResource("").getPath();

Somehow the best answer doesn't work for me. I need to use a slightly different code instead.
ClassLoader loader = Thread.currentThread().getContextClassLoader();
InputStream is = loader.getResourceAsStream("SomeTextFile.txt");
I hope this help those who encounters the same issue.

If you use Guava:
import com.google.common.io.Resources;
we can get URL from CLASSPATH:
URL resource = Resources.getResource("test.txt");
String file = resource.getFile(); // get file path
or InputStream:
InputStream is = Resources.getResource("test.txt").openStream();
Ways to convert an InputStream to a String

To read the contents of a file into a String from the classpath, you can use this:
private String resourceToString(String filePath) throws IOException, URISyntaxException
{
try (InputStream inputStream = this.getClass().getClassLoader().getResourceAsStream(filePath))
{
return IOUtils.toString(inputStream);
}
}
Note:
IOUtils is part of Commons IO.
Call it like this:
String fileContents = resourceToString("ImOnTheClasspath.txt");

You say "I am trying to read a text file which is set in CLASSPATH system variable." My guess this is on Windows and you are using this ugly dialog to edit the "System Variables".
Now you run your Java program in the console. And that doesn't work: The console gets a copy of the values of the system variables once when it is started. This means any change in the dialog afterwards doesn't have any effect.
There are these solutions:
Start a new console after every change
Use set CLASSPATH=... in the console to set the copy of the variable in the console and when your code works, paste the last value into the variable dialog.
Put the call to Java into .BAT file and double click it. This will create a new console every time (thus copying the current value of the system variable).
BEWARE: If you also have a User variable CLASSPATH then it will shadow your system variable. That is why it is usually better to put the call to your Java program into a .BAT file and set the classpath in there (using set CLASSPATH=) rather than relying on a global system or user variable.
This also makes sure that you can have more than one Java program working on your computer because they are bound to have different classpaths.

My answer is not exactly what is asked in the question. Rather I am giving a solution exactly how easily we can read a file into out java application from our project class path.
For example suppose a config file name example.xml is located in a path like below:-
com.myproject.config.dev
and our java executable class file is in the below path:-
com.myproject.server.main
now just check in both the above path which is the nearest common directory/folder from where you can access both dev and main directory/folder (com.myproject.server.main - where our application’s java executable class is existed) – We can see that it is myproject folder/directory which is the nearest common directory/folder from where we can access our example.xml file. Therefore from a java executable class resides in folder/directory main we have to go back two steps like ../../ to access myproject. Now following this, see how we can read the file:-
package com.myproject.server.main;
class Example {
File xmlFile;
public Example(){
String filePath = this.getClass().getResource("../../config/dev/example.xml").getPath();
this.xmlFile = new File(filePath);
}
public File getXMLFile() {
return this.xmlFile;
}
public static void main(String args[]){
Example ex = new Example();
File xmlFile = ex.getXMLFile();
}
}

If you compile your project in jar file:
you can put your file in resources/files/your_file.text or pdf;
and use this code:
import org.slf4j.Logger;
import org.slf4j.LoggerFactory;
import java.io.*;
public class readFileService(){
private static final Logger LOGGER = LoggerFactory.getLogger(readFileService.class);
public byte[] getFile(){
String filePath="/files/your_file";
InputStream inputStreamFile;
byte[] bytes;
try{
inputStreamFile = this.getClass().getResourceAsStream(filePath);
bytes = new byte[inputStreamFile.available()];
inputStreamFile.read(bytes);
} catch(NullPointerException | IOException e) {
LOGGER.error("Erreur read file "+filePath+" error message :" +e.getMessage());
return null;
}
return bytes;
}
}

I am using webshpere application server and my Web Module is build on Spring MVC. The Test.properties were located in the resources folder, i tried to load this files using the following:
this.getClass().getClassLoader().getResourceAsStream("Test.properties");
this.getClass().getResourceAsStream("/Test.properties");
None of the above code loaded the file.
But with the help of below code the property file was loaded successfully:
Thread.currentThread().getContextClassLoader().getResourceAsStream("Test.properties");
Thanks to the user "user1695166".

Use org.apache.commons.io.FileUtils.readFileToString(new File("src/test/resources/sample-data/fileName.txt"));

Don't use getClassLoader() method and use the "/" before the file name. "/" is very important
this.getClass().getResourceAsStream("/SomeTextFile.txt");

import java.io.BufferedReader;
import java.io.File;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
public class ReadFile
{
/**
* * feel free to make any modification I have have been here so I feel you
* * * #param args * #throws InterruptedException
*/
public static void main(String[] args) throws InterruptedException {
// thread pool of 10
File dir = new File(".");
// read file from same directory as source //
if (dir.isDirectory()) {
File[] files = dir.listFiles();
for (File file : files) {
// if you wanna read file name with txt files
if (file.getName().contains("txt")) {
System.out.println(file.getName());
}
// if you want to open text file and read each line then
if (file.getName().contains("txt")) {
try {
// FileReader reads text files in the default encoding.
FileReader fileReader = new FileReader(
file.getAbsolutePath());
// Always wrap FileReader in BufferedReader.
BufferedReader bufferedReader = new BufferedReader(
fileReader);
String line;
// get file details and get info you need.
while ((line = bufferedReader.readLine()) != null) {
System.out.println(line);
// here you can say...
// System.out.println(line.substring(0, 10)); this
// prints from 0 to 10 indext
}
} catch (FileNotFoundException ex) {
System.out.println("Unable to open file '"
+ file.getName() + "'");
} catch (IOException ex) {
System.out.println("Error reading file '"
+ file.getName() + "'");
// Or we could just do this:
ex.printStackTrace();
}
}
}
}
}
}

you have to put your 'system variable' on the java classpath.