Develop Reference

How to get the path of a file inside a project to read it with OpenCSV

I have a CSV file inside a folder that's inside the source folder but I can't get to it.
I got it to work with what I've found on internet:
URL url = getClass().getResource("/csv/recetas.csv");
File file = new File(url.getPath());
FileReader fileReader = new FileReader(file);
CSVReader csvReader = new CSVReader(fileReader, ',', '"', 1);
but it only works when I run it in the IDE. When I build the jar and try to run it, the FileReader can't find the file, it doesn't throw error for URL or File.
Here is my project folder so you can understand me. Thanks.

InputStream in = getClass().getResourceAsStream("/csv/recetas.csv");
InputStreamReader reader = new InputStreamReader(in, StandardCharsets.UTF_8);
CSVReader(reader, ',', '"', 1);
Resources are class path "files" possibly packed in a jar. They are not File, and are read-only.
Also for compatibility, give the charset explicitly.

The getClass().getResource() uses the class loader to load the resource. This means that your csv file will not be visible unless it is in the classpath.
Looking at your code and your problem again, getClass().getResource() seems redundant to me as the constructor of File(...) accepts a file path depicted as String.
See:
https://docs.oracle.com/javase/7/docs/api/java/io/File.html
Quote:
File(String pathname) Creates a new File instance by converting the
given pathname string into an abstract pathname.
To make your program more versatile I would recommend you to avoid hardcoding the file path as the csv file can be anywhere within the file system and it may not always be called recetas.csv.
What people typically would do is to make the java program accept an option like --csv. Then let the user specify the filepath and your code will just be new File(theSpecifiedPath).

How can I package a resource file into a runnable jar

I have tried this in many ways, and followed several online instructions, such as this -
Eclipse exported Runnable JAR not showing images
But so far, no success.
I have a very simple program that reads a text file (test.txt) and prints out its contents to standard output. I am using Eclipse. Here is what I have done-
I put the text file into a folder called 'resources'
I right-clicked the resources folder, selected "Build path" and hit "Use as resource folder"
In my code, I tried two ways of addressing the file - (a) as "/test.txt" and (b) as "resources/test.txt. The latter method works when I run the program within Eclipse. However, neither method works when I export the project as a runnable jar file. The resource folder fails to be included in the jar file.
Any help will be much appreciated.
Addendum:
Here is the code.
public class FileIOTest {
public static void main(String[] args) {
String line;
try {
BufferedReader br = new BufferedReader(new FileReader(new File("/test.txt")));
while((line = br.readLine()) != null){
System.out.println(line);
}
br.close();
}
catch (FileNotFoundException e) {e.printStackTrace();} catch (IOException e) {e.printStackTrace();}
}
}

You can not access a resource in a Jar file through a FileReader because it is not a file on the file system. Java does have a mechanism for accessing resources in your classpath (including in Jars): ClassLoader.getResourceAsStream(). You can use it with your default classloader:
BufferedReader br = new BufferedReader(new InputStreamReader(
ClassLoader.getSystemClassLoader()
.getResourceAsStream("test.txt")));
Note that getResourceAsStream returns null instead of throwing an exception, so you may want to get the stream first, check if for null, then either throw an exception or process it.
You will want to give the full path within the jar as your path.

Files work differently from resources. If you want a resource, then you should not be trying to use any of the File(Xxx) variants, which will look on the file system.
The link you provided, used Class.getResource which returns a URL, but there is also Class.getResourceAsStream which returns an InputStream, which you can wrap in an InputStreamReader. For example
InputStream is = FileIOTest.class.getResourceAsStream("/text.txt");
BufferedReader reader = new BufferedReader(new InputStreamReader(is));

You can create jar without eclipse. For example, if I have a Demo class in package src (directory is the same), and the resource file (named testing.txt) is in resources directory. I add a Manifest.txt file # the root of the directory :
Manifest-Version: 1.0
Main-Class: src.Demo
The Demo class looks like :
package src;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class Demo {
public static void main(String... args) {
try (BufferedReader br = new BufferedReader(new InputStreamReader(Demo.class.getResourceAsStream("/resources/testing.txt"))))
{
String sCurrentLine;
while ((sCurrentLine = br.readLine()) != null) {
System.out.println(sCurrentLine);
}
} catch (IOException e) {
e.printStackTrace();
}
}
}
The testing.txt file is like :
hello world
How are you?
Compile Demo.java to obtain Demo.class
javac src/Demo.java
Create jar
jar cfm myJar.jar Manifest.txt src/* resources/*
And run it
java -jar myJar.jar
In eclipse, just add the resources folder at the classpath (be sure it is in the jar), and read it as a classpath resource (the getResourceAsStream method)

Create File at user defined path

I am creating a file in java using
BufferedWriter out = new BufferedWriter(new FileWriter(FileName));
StringBuffer sb=new StringBuffer();
sb.append("\n");
sb.append("work");
out.write(sb.toString());
out.close();
But this file is getting created inside the bin folder of my server.I would like to create this file inside a user-defined folder.
How can it be achieved.

I would like to create this file inside a user-defined folder.
The simplest approach is to specify a fully qualified path name. You could select that as a File and build a new File relative to it:
File directory = new File("/home/jon/somewhere");
File fullPath = new File(directory, fileName);
BufferedWriter writer = new BufferedWriter(
new OutputStreamWriter(
(new FileOutputStream(fullPath), charSet));
try {
writer.write("\n");
writer.write("work");
} finally {
writer.close();
}
Note:
I would suggest using a FileOutputStream wrapped in an OutputStreamWriter instead of using FileWriter, as you can't specify an encoding with FileWriter
Use a try/finally block (or try-with-resources in Java 7) so that you always close the writer even if there's an exception.

To create a file in a specific directory, you need to specify it in the file name.
Otherwise it will use the current working directory which is likely to be where the program was started from.
BTW: Unless you are using Java 1.4 or older, you can use StringBuilder instead of StringBuffer, although in this case PrintWriter would be even better.

java reading text from src without writing a long path

I can read texts and write them to console however when i install this application to another computer wherever it is installed I dont want to change the path of the txt file. I want to write it like
BufferedReader in = new BufferedReader(new FileReader("xxx.txt"));
I don't want to:
BufferedReader in = new BufferedReader(new FileReader("C:\\Users\\abcde\\Desktop\\xxx.txt"));
is there any way to show this txt file? By the way I put this txt file inside the sources but it cant read!

First get the default application path then check if file exist if exist continue if not close application.
String path = System.getProperty("user.dir");
System.out.println(path + "\\disSoruCevap.txt");
File file = new File(path + "\\disSoruCevap.txt");
if (!file.exists()) {
System.out.println("System couldnt file source file!");
System.out.println("Application will explode");
}
EDIT*
Please prefer one of the answer using resource streams, as you will
see from comments using user.dir is not safe in every case.

You are looking for :
BufferedReader in = new BufferedReader(getClass().getResourceAsStream("/xxx.txt"));
This will load xxx.txt from your jar file (or any jar file in your class path that has that file inside its root directory).

URL fileURL= yourClassName.class.getResource("yourFileName.extension");
String myURL= fileURL.toString();
now you don't need long path name PLUS this one is dynamic in nature i.e., you can now move your project to any pc, any drive.This is because it access URL by using your CLASS location not by any static location (like c:\folder\ab.mp3, then you can't access that file if you move to D drive because then you have to change to D:/folder/ab.mp3 manually which is static in nature)(NOTE: just keep that file with your project)
You can use fileURL as: File file=new File(fileURL.toURI());
You can use myURL as: Media musicFile=new Media(myURL); //in javaFX which need string not url of file

InputStream input = Class_name.class.getResourceAsStream("/xxx.txt");
InputStreamReader inputReader = new InputStreamReader(input);
BufferedReader br = new BufferedReader(inputReader);
String line = null;
try {
while((line = br.readLine())!=null){
System.out.println(line);
}
} catch (IOException ex) {
ex.printStackTrace();
}
You don't need to write or mention long path. Using this code Class_name.class.getResourceAsStream("/xxx.txt"), you can easily get your file.

BufferedReader in = new BufferedReader(new FileReader("xxx.txt")); works fine because when you run your application on an IDE, xxx.txt apparantly is lying in Java's working directory.
Working directory is an operating system feature and it can not be changed.
There are a few ways to deal with this.
1 - use file constructor new File(parent, filename); and load parent using a public static final constant or a property (either passed from command line or otherwise)
2 - or use InputStream in = YourClass.class.getClassLoader().getResourceAsStream("xxx.txt"); - provided your xxx.txt file is packaged under same location as YourClass

Try:
InputStream is = ClassLoader.getSystemResourceAsStream("xxx.txt");
BufferedReader in = new BufferedReader(new InputStreamReader(is));
Depending on where exactly is your file compared to the root of your classpath, you may have to replace xxx.txt3 with /xxx.txt.

My file paths are like this:
public final static String COURSE_FILE_LOCATION = "src/main/resources/courses.csv";
public final static String PREREQUISITE_FILE_LOCATION = "src/main/resources/prerequisites.csv";
This doesn't work. So I delete the .iml file, .idea and target folder from the project and reload them.
Read the correct path like this:
This would work then.

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);
}