I have been trying to read a file into Eclipse. I've looked over other questions, but those answers did not remedy the situation (refreshing the project folder, using getProperty and specifying the correct path, etc.) I've moved the file into every folder and I get the same error. I've copied the file into the directory as shown here:
I've also pasted the code below. It's stupidly simple. The error I get is "FileInputStream.open(String) line: not available [native method]".
Any help would be appreciated. Code is below.
import java.io.File;
import java.io.FileNotFoundException;
import java.util.Scanner;
import java.util.StringTokenizer;
public class Modulo {
public static void main(String[] args) throws FileNotFoundException {
File questions = new File("input.txt");
Scanner sc = new Scanner(questions);
while(sc.hasNext()){
int x = sc.nextInt();
String divide = sc.next();
int y = sc.nextInt();
System.out.println(x % y);
}
}
}
The answer depends.
If you want the file to be embedded within your application when your deploy it (as a Jar file for example), then you can't use File to reference it, as you've tried to include it within your application.
Eclipse further complicates the matter, as you can't included resources within your src directory, but needs to be maintained within a resources directory at the same level as your src folder (this folder may need to be included as part of your build process, but I only have a passing knowledge of how Eclipse works)...
Once you've corrected for all this, you will then need to use Class#getResource to load the resource...for example...
try (InputStream is = Modulo.class.getResourceAsStream("/input.txt")) {
Scanner sc = new Scanner(is);
//...
} catch (IOException exp) {
exp.printStackTrace();
}
However, if you want the file to be an external resource to your program, then you need to place it within a location relative to the location that the program is executed.
Normally, I would suggest the project directory, but I have a funny feeling that Eclipse run's it's Java programs in a different location ... and I don't know if you can change it...
In this case, you could use System.out.println(new File(".").getAbsolutePath()); or System.out.println(new File(".").getCanonicalPath()); or System.out.println(System.getProperty("user.dir")); which will tell you where you program is currently running and place the file there.
Of course, once build (into a Jar) you would need to place the file within a context that was relative to the location it was executed from...
You copy that file into Project folder parallel to src. This is the place base path of your code.
Eclipse by default looks for the file in the main project directory in your case Remainders, if your file is not there, you get an exception. Try placing the file directly under your project and run the same program, it should run correctly.
Related
I am attempting to read from a file, however the console gives me this error.
Exception in thread "main" java.io.FileNotFoundException: dataEx.txt (The system cannot find the file specified)
This is the code that I am executing.
import java.io.*;
import java.util.*;
public class ReadTest {
public static void main(String[] args) throws IOException {
BufferedReader f = new BufferedReader(new FileReader("dataEx.txt" ));
}
}
This is my project structure
-project
-ReadTest.java
-dataEx.txt
Working directory
Your path is wrong, thus the reader can not find the file. Whereever you think your current working directory should be, that is not where it is.
Execute the following code to know where it is:
System.out.println(Paths.get("").toAbsolutePath());
That is the path to your current working directory. Then compare that result to your expectation. Realize that your expectation was wrong and correct the path to your file or your working directory settings.
It is hard to guess where your directory might be right now. Maybe in your bin folder, next to the .class files. You will see after executing the above code snippet.
NIO
By the way, not sure what exactly you plan on doing with that BufferedReader but you might be interested in the newer modern file API revolving around Files and Paths:
List<String> lines = Files.readAllLines(Paths.get("myFile.txt"));
It also has other neat utility methods for File IO, much better than the cumbersome File class and the clunky BufferedReader.
I have a file named "word.txt".
It is in the same directory as my java file.
But when I try to access it in the following code this file not found error occurs:
Exception in thread "main" java.io.FileNotFoundException: word.txt
(The system cannot find the file specified)
at java.io.FileInputStream.open(Native Method)
at java.io.FileInputStream.<init>(Unknown Source)
at java.util.Scanner.<init>(Unknown Source)
at Hangman1.main(Hangman1.java:6)
Here's my code:
import java.io.File;
import java.util.*;
public class Hangman1 {
public static void main(String[] args) throws Exception {
Scanner input = new Scanner(new File("word.txt"));
String in = "";
in = input.nextLine();
}
}
Put the word.txt directly as a child of the project root folder and a peer of src
Project_Root
src
word.txt
Disclaimer: I'd like to explain why this works for this particular case and why it may not work for others.
Why it works:
When you use File or any of the other FileXxx variants, you are looking for a file on the file system relative to the "working directory". The working directory, can be described as this:
When you run from the command line
C:\EclipseWorkspace\ProjectRoot\bin > java com.mypackage.Hangman1
the working directory is C:\EclipseWorkspace\ProjectRoot\bin. With your IDE (at least all the ones I've worked with), the working directory is the ProjectRoot. So when the file is in the ProjectRoot, then using just the file name as the relative path is valid, because it is at the root of the working directory.
Similarly, if this was your project structure ProjectRoot\src\word.txt, then the path "src/word.txt" would be valid.
Why it May not Work
For one, the working directory could always change. For instance, running the code from the command line like in the example above, the working directory is the bin. So in this case it will fail, as there is not bin\word.txt
Secondly, if you were to export this project into a jar, and the file was configured to be included in the jar, it would also fail, as the path will no longer be valid either.
That being said, you need to determine if the file is to be an embedded-resource (or just "resource" - terms which sometimes I'll use interchangeably). If so, then you will want to build the file into the classpath, and access it via an URL. First thing you would need to do (in this particular) case is make sure that the file get built into the classpath. With the file in the project root, you must configure the build to include the file. But if you put the file in the src or in some directory below, then the default build should put it into the class path.
You can access classpath resource in a number of ways. You can make use of the Class class, which has getResourceXxx method, from which you use to obtain classpath resources.
For example, if you changed your project structure to ProjectRoot\src\resources\word.txt, you could use this:
InputStream is = Hangman1.class.getResourceAsStream("/resources/word.txt");
BufferedReader reader = new BufferedReader(new InputStreamReader(is));
getResourceAsStream returns an InputStream, but obtains an URL under the hood. Alternatively, you could get an URL if that's what you need. getResource() will return an URL
For Maven users, where the directory structure is like src/main/resources, the contents of the resources folder is put at the root of the classpath. So if you have a file in there, then you would only use getResourceAsStream("/thefile.txt")
Relative paths can be used, but they can be tricky. The best solution is to know where your files are being saved, that is, print the folder:
import java.io.File;
import java.util.*;
public class Hangman1 {
public static void main(String[] args) throws Exception {
File myFile = new File("word.txt");
System.out.println("Attempting to read from file in: "+myFile.getCanonicalPath());
Scanner input = new Scanner(myFile);
String in = "";
in = input.nextLine();
}
}
This code should print the folder where it is looking for. Place the file there and you'll be good to go.
Your file should directly be under the project folder, and not inside any other sub-folder.
If the folder of your project is named for e.g. AProject, it should be in the same place as your src folder.
Aproject
src
word.txt
Try to create a file using the code, so you will get to know the path of the file where the system create
File test=new File("check.txt");
if (test.createNewFile()) {
System.out.println("File created: " + test.getName());
}
I was reading path from a properties file and didn't mention there was a space in the end.
Make sure you don't have one.
Make sure when you create a txt file you don't type in the name "name.txt", just type in "name". If you type "name.txt" Eclipse will see it as "name.txt.txt". This solved it for me. Also save the file in the src folder, not the folder were the .java resides, one folder up.
I have the same problem, but you know why? because I didn't put .txt in the end of my File and so it was File not a textFile, you shoud do just two things:
Put your Text File in the Root Directory (e.x if you have a project called HelloWorld, just right-click on the HelloWorld file in the package Directory and create File
Save as that File with any name that you want but with a .txt in the end of that
I guess your problem is solved, but I write it to other peoples know that.
Thanks.
i think it always boils to the classpath. having said that if you run from the same folder where your .class is then change Scanner input = new Scanner(new File("word.txt")); to Scanner input = new Scanner(new File("./word.txt")); that should work
In Python the global variable __file__ is the full path of the current file.
System.getProperty("user.dir"); seems to return the path of the current working directory.
I want to get the path of the current .java, .class or package file.
Then use this to get the path to an image.
My project file structure in Netbeans looks like this:
(source: toile-libre.org)
Update to use code suggested from my chosen best answer:
// read image data from picture in package
try {
InputStream instream = TesseractTest.class
.getResourceAsStream("eurotext.tif");
bufferedImage = ImageIO.read(instream);
}
catch (IOException e) {
System.out.println(e.getMessage());
}
This code is used in the usage example from tess4j.
My full code of the usage example is here.
If you want to load an image file stored right next to your class file, use Class::getResourceAsStream(String name).
In your case, that would be:
try (InputStream instream = TesseractTest.class.getResourceAsStream("eurotext.tif")) {
// read stream here
}
This assumes that your build system copies the .tif file to your build folder, which is commonly done by IDEs, but requires extra setup in build tools like Ant and Gradle.
If you package your program to a .jar file, the code will still work, again assuming your build system package the .tif file next to the .class file.
Is there a way to get the file path of the .java file executed or compiled?
For completeness, the literal answer to your question is "not easily and not always".
There is a round-about way to find the source filename for a class on the callstack via StackFrameElement.getFileName(). However, the filename won't always be available1 and it won't necessarily be correct2.
Indeed, it is quite likely that the source tree won't be present on the system where you are executing the code. So if you needed an image file that was stashed in the source tree, you would be out of luck.
1 - It depends on the Java compiler and compilation options that you use. And potentially on other things.
2 - For example, the source tree can be moved or removed after compilation.
Andreas has described the correct way to solve your problem. Make sure that the image file is in your application's JAR file, and access it using getResource or getResourceAsStream. If your application is using an API that requires a filename / pathname in the file system, you may need to extract the resource from the JAR to a temporary file, or something like that.
public class Main {
public static void main(String[] args) throws Exception {
System.out.println(getPackageParent(Main.class, false));
}
public static String getPackageParent(Class<?> cls, boolean include_last_dot)
throws Exception {
StringBuilder sb = new StringBuilder(cls.getPackage().getName());
if (sb.lastIndexOf(".") > 0)
if (include_last_dot)
return sb.delete(sb.lastIndexOf(".") + 1, sb.length())
.toString();
else
return sb.delete(sb.lastIndexOf("."), sb.length()).toString();
return sb.toString();
}
}
by running this...
File file = new File("Highscores.scr");
i keep getting this error, and i really don't know how to get around it.
the file is currently sitting in my source packages with my .java files.
I can quite easily read the file by specifying the path but i intend to run this on multiple computers so i need the file to be portable with the program.
this question isnt about reading the text file but rather specifying its location without using an absolute path .
ive searched for the answer but the answers i get are just "specify the name" and "specify the absolute path".
id post an image to make it more clear but i dont have the 10 rep to do so :/
how do i do this?
cheers.
The best way to do this is to put it in your classpath then getResource()
package com.sandbox;
import org.apache.commons.io.FileUtils;
import java.io.File;
import java.io.IOException;
import java.net.URISyntaxException;
import java.net.URL;
public class Sandbox {
public static void main(String[] args) throws URISyntaxException, IOException {
new Sandbox().run();
}
private void run() throws URISyntaxException, IOException {
URL resource = Sandbox.class.getResource("/my.txt");
File file = new File(resource.toURI());
String s = FileUtils.readFileToString(file);
System.out.println(s);
}
}
I'm doing this because I'm assuming you need a File. But if you have an api which takes an InputStream instead, it's probably better to use getResourceAsStream instead.
Notice the path, /my.txt. That means, "get a file named my.txt that is in the root directory of the classpath". I'm sure you can read more about getResource and getResourceAsStream to learn more about how to do this. But the key thing here is that the classpath for the file will be the same for any computer you give the executable to (as long as you don't move the file around in your classpath).
BTW, if you get a null pointer exception on the line that does new File, that means that you haven't specified the correct classpath for the file.
As far as I remember the default directory with be the same as your project folder level. Put the file one level higher.
-Project/
----src/
----test/
-Highscores.scr
If you are building your code on your eclipse then you need to put your Highscores.scr to your project folder. Try that and check.
You can try to run the following sample program to check which is the current directory your program is picking up.
File f = new File(".");
System.out.println("Current Directory is: " + f.getAbsolutePath());
I am reading a file as follows:
File imgLoc = new File("Player.gif");
BufferedImage image = null;
try {
image = ImageIO.read(imgLoc);
}
catch(Exception ex)
{
System.out.println("Image read error");
System.exit(1);
}
return image;
I do not know where to place my file to make the Eclipse IDE, and my project can detect it when I run my code.
Is there a better way of creating a BufferedImage from an image file stored in your project directory?
Take a look in the comments for Class.getResource and Class.getResourceAsStream. These are probably what you really want to use as they will work whether you are running from within the directory of an Eclipse project, or from a JAR file after you package everything up.
You use them along the lines of:
InputStream in = MyClass.class.getResourceAsStream("Player.gif");
In this case, Java would look for the file "Player.gif" next to the MyClass.class file. That is, if the full package/class name is "com.package.MyClass", then Java will look for a file in "[project]/bin/com/package/Player.gif". The comments for getResourceAsStream indicate that if you lead with a slash, i.e. "/Player.gif", then it'll look in the root (i.e. the "bin" directory).
Note that you can drop the file in the "src" directory and Eclipse will automatically copy it to the "bin" directory at build time.
In the run dialog you can choose the directory. The default is the project root.
From my experience it seems to be the containing projects directory by default, but there is a simple way to find out:
System.out.println(new File(".").getAbsolutePath());
Are you trying to write a plugin for Eclipse or is it a regular project?
In the latter case, wouldn't that depend on where the program is installed and executed in the end?
While trying it out and running it from Eclipse, I'd guess that it would find the file in the project workspace. You should be able to find that out by opening the properties dialog for the project, and looking under the Resource entry.
Also, you can add resources to a project by using the Import menu option.
The default root folder for any Eclipse project is also a relative path of that application.
Below are steps I used for my Eclipse 4.8.0 and Java 1.8 project.
I - Place your file you want to interact with along the BIN and SRS folders of your project and not in one of those folders.
II - Implement below code in your main() method.
public static void main(String [] args) throws IOException {
FileReader myFileReader;
BufferedReader myReaderHelper;
try {
String localDir = System.getProperty("user.dir");
myFileReader = new FileReader(localDir + "\\yourFile.fileExtension");
myReaderHelper = new BufferedReader(myFileReader);
if (myReaderHelper.readLine() != null) {
StringTokenizer myTokens =
new StringTokenizer((String)myReaderHelper.readLine(), "," );
System.out.println(myTokens.nextToken().toString()); // - reading first item
}
} catch (FileNotFoundException myFileException) {
myFileException.printStackTrace(); } } // End of main()
III - Implement a loop to iterate through elements of your file if your logic requires this.