Develop Reference

Java.io.FileNotFoundException when specifying a directory

I'm trying to define a File in Java with a txt file called "helloworld". I've placed this file in a resources folder and when making the file I defined it like this:
File file = new File("/helloworld");
However I get this error when compiling
Exception in thread "main" java.io.FileNotFoundException: /helloworld
(No such file or directory)
at java.io.FileInputStream.open0(Native Method)
at java.io.FileInputStream.open(FileInputStream.java:195)
at java.io.FileInputStream.<init>(FileInputStream.java:138)
at java.io.FileReader.<init>(FileReader.java:72)
at Tests.main(Tests.java:15)
This is the entire code I am trying to execute if that helps troubleshoot this issue
// Java Program to illustrate reading from FileReader
// using BufferedReader
import java.io.BufferedReader;
import java.io.File;
import java.io.FileReader;
import java.net.URL;
public class Tests
{
public static void main(String[] args)throws Exception
{
File file = new File("/helloworld");
BufferedReader br = new BufferedReader(new FileReader(file));
String st;
while ((st = br.readLine()) != null)
System.out.println(st);
}
}
Thank you for the help!

public File​(String pathname)
Creates a new File instance by converting the given pathname string
into an abstract pathname. If the given string is the empty string,
then the result is the empty abstract pathname.
You are trying to create a new File instance but the file called helloworld is not found or some other reasons. That's why you get the error,
Exception in thread "main" java.io.FileNotFoundException: /helloworld
The named file does not exist.
The named file is actually a directory.
The named file cannot be opened for reading for some reason.
You say that you try to define a file but your code seems to read. Try below one if you want to create a file,
import java.io.*;
import java.nio.charset.StandardCharsets;
class TestDir {
public static void main(String[] args) {
String fileName = "filename.txt";
try (Writer writer = new BufferedWriter(new OutputStreamWriter(
new FileOutputStream(fileName), StandardCharsets.UTF_8))) {
writer.write("write something in text file");
} catch (IOException e) {
e.printStackTrace();
}
}
}

It is easy to diagnose: The path you specified starts with a slash, so it means that the file is expected to be located at the root directory of the filesystem. You'd better strip off the leading slash, and:
Either start your program at the same directory the file is at.
Either specify an absolute/relative path in your code when instantiating the File object.

If the file is in a resources folder and is intended to be bundled with your program, you need to treat it like a resource, not a file.
This means you should not use the File class. You should read your data with the Class.getResource or Class.getResourceAsStream method:
BufferedReader br = new BufferedReader(
new InputStreamReader(
Tests.class.getResourceAsStream("/helloworld")));
This becomes especially important if you want to distribute a program as a .jar file. A .jar file is a compressed archive (actually a zip file with different extension) which contains both compiled class files and resources. Since they are all compressed into one .jar file, they are not individual files at all, so there is no way the File class can refer to them.
Although the File class is not useful for what you’re trying to do, you may want to research the concept of absolute file names and relative file names. By starting a file name with /, you are specifying an absolute file name, which means you are telling the program to look for the file in a specific place—a place where the file almost certain will not reside.

Try bellow to know where is the folder or file's path, which your program is looking for
System.out.println(file.getAbsolutePath());
With
File file = new File("/helloworld");
I think your program is looking for c:\helloworld, and there is no file or folder's name is helloword in your C drive
If you put the helloword.txt into C drive, and
File file = new File("C:\\helloworld.txt");
FileNotFoundException will disappear.

Relative path to store a file in any computer - Java

I have used this reference to read a file on my project. but i need to store likewise i don't know how to do this? please help me.
for reading a file i have taken the code from the link
InputStream csv =
SomeClassInTheSamePackage.class.getResourceAsStream("filename.csv");
Like this can anyone help me with writing a file
currently I'm using this code:
Writer output = null;
output = new BufferedWriter(new FileWriter("./filename.csv"));
But it throws FileNotFound Exception at runtime
Issue is with locating file path. it works fine if i give absolute path. but it needs to be run in any computer

The FileWriter will create files as required. It doesn't have to exist already.
The only way you can get this error is if the current working directory no longer exists or you don't have permission to create a file in the directory.

Check your permissions:
Writer output = null;
if (new File("./filename.csv").canWrite()) {
System.out.println("You have not permissions");
} else {
output = new BufferedWriter(new FileWriter("./filename.csv"));
...
}

OK..Use the following code to get the path of file:
String path = System.getProperty("user.dir");
FileWriter fw = new FileWriter(path+File.separator+"filename.csv");
EDIT
Here is a Demo Code that is creating file "pqr.txt" in package myPackage. Given that I am executing the class file using command java myPackage.AccessCheck i.e from one directory above the mypackage.
package myPackage;
import java.io.FileWriter;
import java.io.File;
public class AccessCheck
{
protected void callme()
{
try
{
String name = System.getProperty("user.dir")+File.separator+AccessCheck.class.getPackage().getName()+File.separator+"pqr.txt";
System.out.println(name);
FileWriter fw = new FileWriter(name);
}catch(Exception ex){ex.printStackTrace();}
}
public static void main(String st[])
{
AccessCheck asc = new AccessCheck();
asc.callme();
}
}

Instead of calling getResourceAsStream(), call class.getResource() or classLoader.getResource(). These functions return a URL giving the location of the resource. If the resource is a plain file on the filesystem, the URL will be a file: URL.
Once you have the URL, you can convert it to a URI and pass it to the File constructor which takes a URI. Then you can use the File object to open the file.
Note that getResource() can return URLs that the File constructor can't deal with. For example, if the resource is found in a jar file, then the URL might look like this:
jar:file:/C:/path/to/JDK/lib/vm.jar!/java/lang/Object.class
The File constructor can't deal with URLs like this. This may or may not be a concern for you.

java.io.FileNotFoundException (File not found) using Scanner. What's wrong in my code?

I've a .txt file ("file.txt") in my netbeans "/build/classes" directory.
In the same directory there is the .class file compiled for the following code:
try {
File f = new File("file.txt");
Scanner sc = new Scanner(f);
}
catch (IOException e) {
System.out.println(e);
}
Debugging the code (breakpoint in "Scanner sc ..") an exception is launched and the following is printed:
java.io.FileNotFoundException: file.txt (the system can't find the
specified file)
I also tried using "/file.txt" and "//file.txt" but same result.
Thank you in advance for any hint

If you just use new File("pathtofile") that path is relative to your current working directory, which is not at all necessarily where your class files are.
If you are sure that the file is somewhere on your classpath, you could use the following pattern instead:
URL path = ClassLoader.getSystemResource("file.txt");
if(path==null) {
//The file was not found, insert error handling here
}
File f = new File(path.toURI());

The JVM will look for the file in the current working directory.
Where this is depends on your IDE settings (how your program is executed).
To figure out where it expects file.txt to be located, you could do
System.out.println(new File("."));
If it for instance outputs
/some/path/project/build
you should place file.txt in the build directory (or specify the proper path relative to the build directory).

Try:
File f = new File("./build/classes/file.txt");

Use "." to denote the current directory
String path = "./build/classes/file.txt";
File f = new File(path);

File Object loads, looking for match in its current directory.... which is Directly in Your project folder where your class files are loaded not in your source ..... put the file directly in the project folder

How to read file from relative path in Java project? java.io.File cannot find the path specified

I have a project with 2 packages:
tkorg.idrs.core.searchengines
tkorg.idrs.core.searchengines
In package (2) I have a text file ListStopWords.txt, in package (1) I have a class FileLoadder. Here is code in FileLoader:
File file = new File("properties\\files\\ListStopWords.txt");
But I have this error:
The system cannot find the path specified
Can you give a solution to fix it?

If it's already in the classpath, then just obtain it from the classpath instead of from the disk file system. Don't fiddle with relative paths in java.io.File. They are dependent on the current working directory over which you have totally no control from inside the Java code.
Assuming that ListStopWords.txt is in the same package as your FileLoader class, then do:
URL url = getClass().getResource("ListStopWords.txt");
File file = new File(url.getPath());
Or if all you're ultimately after is actually an InputStream of it:
InputStream input = getClass().getResourceAsStream("ListStopWords.txt");
This is certainly preferred over creating a new File() because the url may not necessarily represent a disk file system path, but it could also represent virtual file system path (which may happen when the JAR is expanded into memory instead of into a temp folder on disk file system) or even a network path which are both not per definition digestable by File constructor.
If the file is -as the package name hints- is actually a fullworthy properties file (containing key=value lines) with just the "wrong" extension, then you could feed the InputStream immediately to the load() method.
Properties properties = new Properties();
properties.load(getClass().getResourceAsStream("ListStopWords.txt"));
Note: when you're trying to access it from inside static context, then use FileLoader.class (or whatever YourClass.class) instead of getClass() in above examples.

The relative path works in Java using the . specifier.
. means same folder as the currently running context.
.. means the parent folder of the currently running context.
So the question is how do you know the path where the Java is currently looking?
Do a small experiment
File directory = new File("./");
System.out.println(directory.getAbsolutePath());
Observe the output, you will come to know the current directory where Java is looking. From there, simply use the ./ specifier to locate your file.
For example if the output is
G:\JAVA8Ws\MyProject\content.
and your file is present in the folder "MyProject" simply use
File resourceFile = new File("../myFile.txt");
Hope this helps.

The following line can be used if we want to specify the relative path of the file.
File file = new File("./properties/files/ListStopWords.txt");

InputStream in = FileLoader.class.getResourceAsStream("<relative path from this class to the file to be read>");
try {
BufferedReader reader = new BufferedReader(new InputStreamReader(in));
String line = null;
while ((line = reader.readLine()) != null) {
System.out.println(line);
}
} catch (Exception e) {
e.printStackTrace();
}

try .\properties\files\ListStopWords.txt

I could have commented but I have less rep for that.
Samrat's answer did the job for me. It's better to see the current directory path through the following code.
File directory = new File("./");
System.out.println(directory.getAbsolutePath());
I simply used it to rectify an issue I was facing in my project. Be sure to use ./ to back to the parent directory of the current directory.
./test/conf/appProperties/keystore

While the answer provided by BalusC works for this case, it will break when the file path contains spaces because in a URL, these are being converted to %20 which is not a valid file name. If you construct the File object using a URI rather than a String, whitespaces will be handled correctly:
URL url = getClass().getResource("ListStopWords.txt");
File file = new File(url.toURI());

Assuming you want to read from resources directory in FileSystem class.
String file = "dummy.txt";
var path = Paths.get("src/com/company/fs/resources/", file);
System.out.println(path);
System.out.println(Files.readString(path));
Note: Leading . is not needed.

I wanted to parse 'command.json' inside src/main//js/Simulator.java. For that I copied json file in src folder and gave the absolute path like this :
Object obj = parser.parse(new FileReader("./src/command.json"));

For me actually the problem is the File object's class path is from <project folder path> or ./src, so use File file = new File("./src/xxx.txt"); solved my problem

For me it worked with -
String token = "";
File fileName = new File("filename.txt").getAbsoluteFile();
Scanner inFile = null;
try {
inFile = new Scanner(fileName);
} catch (FileNotFoundException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
while( inFile.hasNext() )
{
String temp = inFile.next( );
token = token + temp;
}
inFile.close();
System.out.println("file contents" +token);

If text file is not being read, try using a more closer absolute path (if you wish
you could use complete absolute path,) like this:
FileInputStream fin=new FileInputStream("\\Dash\\src\\RS\\Test.txt");
assume that the absolute path is:
C:\\Folder1\\Folder2\\Dash\\src\\RS\\Test.txt

String basePath = new File("myFile.txt").getAbsolutePath();
this basepath you can use as the correct path of your file

if you want to load property file from resources folder which is available inside src folder, use this
String resourceFile = "resources/db.properties";
InputStream resourceStream = ClassLoader.getSystemClassLoader().getResourceAsStream(resourceFile);
Properties p=new Properties();
p.load(resourceStream);
System.out.println(p.getProperty("db"));
db.properties files contains key and value db=sybase

If you are trying to call getClass() from Static method or static block, the you can do the following way.
You can call getClass() on the Properties object you are loading into.
public static Properties pathProperties = null;
static {
pathProperties = new Properties();
String pathPropertiesFile = "/file.xml";
// Now go for getClass() method
InputStream paths = pathProperties.getClass().getResourceAsStream(pathPropertiesFile);
}

How to pass a text file as a argument?

Im trying to write a program to read a text file through args but when i run it, it always says the file can't be found even though i placed it inside the same folder as the main.java that im running.
Does anyone know the solution to my problem or a better way of reading a text file?

Do not use relative paths in java.io.File.
It will become relative to the current working directory which is dependent on the way how you run the application which in turn is not controllable from inside your application. It will only lead to portability trouble. If you run it from inside Eclipse, the path will be relative to /path/to/eclipse/workspace/projectname. If you run it from inside command console, it will be relative to currently opened folder (even though when you run the code by absolute path!). If you run it by doubleclicking the JAR, it will be relative to the root folder of the JAR. If you run it in a webserver, it will be relative to the /path/to/webserver/binaries. Etcetera.
Always use absolute paths in java.io.File, no excuses.
For best portability and less headache with absolute paths, just place the file in a path covered by the runtime classpath (or add its path to the runtime classpath). This way you can get the file by Class#getResource() or its content by Class#getResourceAsStream(). If it's in the same folder (package) as your current class, then it's already in the classpath. To access it, just do:
public MyClass() {
URL url = getClass().getResource("filename.txt");
File file = new File(url.getPath());
InputStream input = new FileInputStream(file);
// ...
}
or
public MyClass() {
InputStream input = getClass().getResourceAsStream("filename.txt");
// ...
}

Try giving an absolute path to the filename.
Also, post the code so that we can see what exactly you're trying.

When you are opening a file with a relative file name in Java (and in general) it opens it relative to the working directory.
you can find the current working directory of your process using
String workindDir = new File(".").getAbsoultePath()
Make sure you are running your program from the correct directory (or change the file name so that it will be relative to where you are running it from).

If you're using Eclipse (or a similar IDE), the problem arises from the fact that your program is run from a few directories above where the actual source is located. Try moving your file up a level or two in the project tree.
Check out this question for more detail.

The simplest solution is to create a new file, then see where the output file is. That is the correct place to put your input file into.

If you put the file and the class working with it under same package can you use this:
Class A {
void readFile (String fileName) {
Url tmp = A.class.getResource (fileName);
// Or Url tmp = this.getClass().getResource (fileName);
File tmpFile = File (tmp);
if (tmpFile.exists())
System.out.print("I found the file.")
}
}
It will help if you read about classloaders.

say I have a text file input.txt which is located on the desktop
and input.txt has the following content
i came
i saw
i left
and below is the java code for reading that text file
public class ReadInputFromTextFile {
public static void main(String[] args) throws Exception
{
File file = new File(
"/Users/viveksingh/desktop/input.txt");
BufferedReader br
= new BufferedReader(new FileReader(file));
String st;
while ((st = br.readLine()) != null)
System.out.println(st);
}
}
output on the console:
i came
i saw
i left