Error reading my csv file in class in Java Application - java

I need to read a sample.csv file from the same class as my main.
Below you can see the structure.
This is my code but when I run it it gave me the error that the sample.csv is not find!
public int convertFile(String csvFileName, String xmlFileName,
String delimiter) {
int rowsCount = -1;
BufferedReader csvReader;
try {
Document newDoc = domBuilder.newDocument();
// Root element
Element rootElement = newDoc.createElement("XMLCreators");
newDoc.appendChild(rootElement);
// Read csv file
csvReader = new BufferedReader(new FileReader(csvFileName));
CSVReader reader = new CSVReader(new FileReader("sample.csv"));
//CSVReader reader = new CSVReader(csvReader);
String[] nextLine;
int line = 0;
List<String> headers = new ArrayList<String>(5);
while ((nextLine = reader.readNext()) != null) {
....
Can anyone suggest me what is the problem?


Without relative or absolute path, it is directly related to the "current" directory. So it does NOT matter so much that the .csv file is in a certain package, but it does matter that when the application is started the "working" directory is exactly the directory that contains the file.
You are using Eclipse, go to see the Run Configuration (Run menu -> Run Configurations ...) used to launch the application, Arguments tab.
you can get path like this.
URL url = YourClassName.class.getResource("sample.csv");
FileReader fileReader = new FileReader(url.getFile());

Related

MyClass.class.getResourceAsStream() not able to find path in jars but works in .class file [duplicate]

My project has the following structure:
/src/main/java/
/src/main/resources/
/src/test/java/
/src/test/resources/
I have a file in /src/test/resources/test.csv and I want to load the file from a unit test in /src/test/java/MyTest.java
I have this code which didn't work. It complains "No such file or directory".
BufferedReader br = new BufferedReader (new FileReader(test.csv))
I also tried this
InputStream is = (InputStream) MyTest.class.getResourcesAsStream(test.csv))
This also doesn't work. It returns null. I am using Maven to build my project.

Try the next:
ClassLoader classloader = Thread.currentThread().getContextClassLoader();
InputStream is = classloader.getResourceAsStream("test.csv");
If the above doesn't work, various projects have been added the following class: ClassLoaderUtil1 (code here).2
Here are some examples of how that class is used:
src\main\java\com\company\test\YourCallingClass.java
src\main\java\com\opensymphony\xwork2\util\ClassLoaderUtil.java
src\main\resources\test.csv
// java.net.URL
URL url = ClassLoaderUtil.getResource("test.csv", YourCallingClass.class);
Path path = Paths.get(url.toURI());
List<String> lines = Files.readAllLines(path, StandardCharsets.UTF_8);
// java.io.InputStream
InputStream inputStream = ClassLoaderUtil.getResourceAsStream("test.csv", YourCallingClass.class);
InputStreamReader streamReader = new InputStreamReader(inputStream, StandardCharsets.UTF_8);
BufferedReader reader = new BufferedReader(streamReader);
for (String line; (line = reader.readLine()) != null;) {
// Process line
}
Notes
See it in The Wayback Machine.
Also in GitHub.

Try:
InputStream is = MyTest.class.getResourceAsStream("/test.csv");
IIRC getResourceAsStream() by default is relative to the class's package.
As #Terran noted, don't forget to add the / at the starting of the filename

Try following codes on Spring project
ClassPathResource resource = new ClassPathResource("fileName");
InputStream inputStream = resource.getInputStream();
Or on non spring project
ClassLoader classLoader = getClass().getClassLoader();
File file = new File(classLoader.getResource("fileName").getFile());
InputStream inputStream = new FileInputStream(file);

Here is one quick solution with the use of Guava:
import com.google.common.base.Charsets;
import com.google.common.io.Resources;
public String readResource(final String fileName, Charset charset) throws IOException {
return Resources.toString(Resources.getResource(fileName), charset);
}
Usage:
String fixture = this.readResource("filename.txt", Charsets.UTF_8)

Non spring project:
String filePath = Objects.requireNonNull(getClass().getClassLoader().getResource("any.json")).getPath();
Stream<String> lines = Files.lines(Paths.get(filePath));
Or
String filePath = Objects.requireNonNull(getClass().getClassLoader().getResource("any.json")).getPath();
InputStream in = new FileInputStream(filePath);
For spring projects, you can also use one line code to get any file under resources folder:
File file = ResourceUtils.getFile(ResourceUtils.CLASSPATH_URL_PREFIX + "any.json");
String content = new String(Files.readAllBytes(file.toPath()));

For java after 1.7
List<String> lines = Files.readAllLines(Paths.get(getClass().getResource("test.csv").toURI()));
Alternatively you can use Spring utils if you are in Spring echosystem
final val file = ResourceUtils.getFile("classpath:json/abcd.json");
To get to more behind the scene, check out following blog
https://todzhang.com/blogs/tech/en/save_resources_to_files

I faced the same issue.
The file was not found by a class loader, which means it was not packed into the artifact (jar). You need to build the project. For example, with maven:
mvn clean package
So the files you added to resources folder will get into maven build and become available to the application.
I would like to keep my answer: it does not explain how to read a file (other answers do explain that), it answers why InputStream or resource was null. Similar answer is here.

ClassLoader loader = Thread.currentThread().getContextClassLoader();
InputStream is = loader.getResourceAsStream("test.csv");
If you use context ClassLoader to find a resource then definitely it will cost application performance.

Now I am illustrating the source code for reading a font from maven created resources directory,
scr/main/resources/calibril.ttf
Font getCalibriLightFont(int fontSize){
Font font = null;
try{
URL fontURL = OneMethod.class.getResource("/calibril.ttf");
InputStream fontStream = fontURL.openStream();
font = new Font(Font.createFont(Font.TRUETYPE_FONT, fontStream).getFamily(), Font.PLAIN, fontSize);
fontStream.close();
}catch(IOException | FontFormatException ief){
font = new Font("Arial", Font.PLAIN, fontSize);
ief.printStackTrace();
}
return font;
}
It worked for me and hope that the entire source code will also help you, Enjoy!

Import the following:
import java.io.IOException;
import java.io.FileNotFoundException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;
import java.util.ArrayList;
The following method returns a file in an ArrayList of Strings:
public ArrayList<String> loadFile(String filename){
ArrayList<String> lines = new ArrayList<String>();
try{
ClassLoader classloader = Thread.currentThread().getContextClassLoader();
InputStream inputStream = classloader.getResourceAsStream(filename);
InputStreamReader streamReader = new InputStreamReader(inputStream, StandardCharsets.UTF_8);
BufferedReader reader = new BufferedReader(streamReader);
for (String line; (line = reader.readLine()) != null;) {
lines.add(line);
}
}catch(FileNotFoundException fnfe){
// process errors
}catch(IOException ioe){
// process errors
}
return lines;
}

getResource() was working fine with the resources files placed in src/main/resources only. To get a file which is at the path other than src/main/resources say src/test/java you need to create it exlicitly.
the following example may help you
import java.io.BufferedReader;
import java.io.FileReader;
import java.io.IOException;
import java.net.URISyntaxException;
import java.net.URL;
public class Main {
public static void main(String[] args) throws URISyntaxException, IOException {
URL location = Main.class.getProtectionDomain().getCodeSource().getLocation();
BufferedReader br = new BufferedReader(new FileReader(location.getPath().toString().replace("/target/classes/", "/src/test/java/youfilename.txt")));
}
}

You can use the com.google.common.io.Resources.getResource to read the url of file and then get the file content using java.nio.file.Files to read the content of file.
URL urlPath = Resources.getResource("src/main/resource");
List<String> multilineContent= Files.readAllLines(Paths.get(urlPath.toURI()));

if you are loading file in static method then
ClassLoader classLoader = getClass().getClassLoader();
this might give you an error.
You can try this
e.g. file you want to load from resources is resources >> Images >> Test.gif
import org.springframework.core.io.ClassPathResource;
import org.springframework.core.io.Resource;
Resource resource = new ClassPathResource("Images/Test.gif");
File file = resource.getFile();

To read the files from src/resources folder then try this :
DataSource fds = new FileDataSource(getFileHandle("images/sample.jpeg"));
public static File getFileHandle(String fileName){
return new File(YourClassName.class.getClassLoader().getResource(fileName).getFile());
}
in case of non static reference:
return new File(getClass().getClassLoader().getResource(fileName).getFile());

Does the code work when not running the Maven-build jar, for example when running from your IDE? If so, make sure the file is actually included in the jar. The resources folder should be included in the pom file, in <build><resources>.

The following class can be used to load a resource from the classpath and also receive a fitting error message in case there's a problem with the given filePath.
import java.io.InputStream;
import java.nio.file.NoSuchFileException;
public class ResourceLoader
{
private String filePath;
public ResourceLoader(String filePath)
{
this.filePath = filePath;
if(filePath.startsWith("/"))
{
throw new IllegalArgumentException("Relative paths may not have a leading slash!");
}
}
public InputStream getResource() throws NoSuchFileException
{
ClassLoader classLoader = this.getClass().getClassLoader();
InputStream inputStream = classLoader.getResourceAsStream(filePath);
if(inputStream == null)
{
throw new NoSuchFileException("Resource file not found. Note that the current directory is the source folder!");
}
return inputStream;
}
}

this.getClass().getClassLoader().getResource("filename").getPath()

My file in the test folder could not be found even though I followed the answers. It got resolved by rebuilding the project. It seems IntelliJ did not recognize the new file automatically. Pretty nasty to find out.

I got it work on both running jar and in IDE by writing as
InputStream schemaStream =
ProductUtil.class.getClassLoader().getResourceAsStream(jsonSchemaPath);
byte[] buffer = new byte[schemaStream.available()];
schemaStream.read(buffer);
File tempFile = File.createTempFile("com/package/schema/testSchema", "json");
tempFile.deleteOnExit();
FileOutputStream out = new FileOutputStream(tempFile);
out.write(buffer);

This worked pretty fine for me :
InputStream in = getClass().getResourceAsStream("/main/resources/xxx.xxx");
InputStreamReader streamReader = new InputStreamReader(in, StandardCharsets.UTF_8);
BufferedReader reader = new BufferedReader(streamReader);
String content = "";
for (String line; (line = reader.readLine()) != null;) {
content += line;
}

I get it to work without any reference to "class" or "ClassLoader".
Let's say we have three scenarios with the location of the file 'example.file' and your working directory (where your app executes) is home/mydocuments/program/projects/myapp:
a)A sub folder descendant to the working directory:
myapp/res/files/example.file
b)A sub folder not descendant to the working directory:
projects/files/example.file
b2)Another sub folder not descendant to the working directory:
program/files/example.file
c)A root folder:
home/mydocuments/files/example.file (Linux; in Windows replace home/ with C:)
1) Get the right path:
a)String path = "res/files/example.file";
b)String path = "../projects/files/example.file"
b2)String path = "../../program/files/example.file"
c)String path = "/home/mydocuments/files/example.file"
Basically, if it is a root folder, start the path name with a leading slash.
If it is a sub folder, no slash must be before the path name. If the sub folder is not descendant to the working directory you have to cd to it using "../". This tells the system to go up one folder.
2) Create a File object by passing the right path:
File file = new File(path);
3) You are now good to go:
BufferedReader br = new BufferedReader(new FileReader(file));

Can't Load my Assets when running my java AWT game jar by RightClicking on MacOS [duplicate]

My project has the following structure:
/src/main/java/
/src/main/resources/
/src/test/java/
/src/test/resources/
I have a file in /src/test/resources/test.csv and I want to load the file from a unit test in /src/test/java/MyTest.java
I have this code which didn't work. It complains "No such file or directory".
BufferedReader br = new BufferedReader (new FileReader(test.csv))
I also tried this
InputStream is = (InputStream) MyTest.class.getResourcesAsStream(test.csv))
This also doesn't work. It returns null. I am using Maven to build my project.

Try the next:
ClassLoader classloader = Thread.currentThread().getContextClassLoader();
InputStream is = classloader.getResourceAsStream("test.csv");
If the above doesn't work, various projects have been added the following class: ClassLoaderUtil1 (code here).2
Here are some examples of how that class is used:
src\main\java\com\company\test\YourCallingClass.java
src\main\java\com\opensymphony\xwork2\util\ClassLoaderUtil.java
src\main\resources\test.csv
// java.net.URL
URL url = ClassLoaderUtil.getResource("test.csv", YourCallingClass.class);
Path path = Paths.get(url.toURI());
List<String> lines = Files.readAllLines(path, StandardCharsets.UTF_8);
// java.io.InputStream
InputStream inputStream = ClassLoaderUtil.getResourceAsStream("test.csv", YourCallingClass.class);
InputStreamReader streamReader = new InputStreamReader(inputStream, StandardCharsets.UTF_8);
BufferedReader reader = new BufferedReader(streamReader);
for (String line; (line = reader.readLine()) != null;) {
// Process line
}
Notes
See it in The Wayback Machine.
Also in GitHub.

Try:
InputStream is = MyTest.class.getResourceAsStream("/test.csv");
IIRC getResourceAsStream() by default is relative to the class's package.
As #Terran noted, don't forget to add the / at the starting of the filename

Try following codes on Spring project
ClassPathResource resource = new ClassPathResource("fileName");
InputStream inputStream = resource.getInputStream();
Or on non spring project
ClassLoader classLoader = getClass().getClassLoader();
File file = new File(classLoader.getResource("fileName").getFile());
InputStream inputStream = new FileInputStream(file);

Here is one quick solution with the use of Guava:
import com.google.common.base.Charsets;
import com.google.common.io.Resources;
public String readResource(final String fileName, Charset charset) throws IOException {
return Resources.toString(Resources.getResource(fileName), charset);
}
Usage:
String fixture = this.readResource("filename.txt", Charsets.UTF_8)

Non spring project:
String filePath = Objects.requireNonNull(getClass().getClassLoader().getResource("any.json")).getPath();
Stream<String> lines = Files.lines(Paths.get(filePath));
Or
String filePath = Objects.requireNonNull(getClass().getClassLoader().getResource("any.json")).getPath();
InputStream in = new FileInputStream(filePath);
For spring projects, you can also use one line code to get any file under resources folder:
File file = ResourceUtils.getFile(ResourceUtils.CLASSPATH_URL_PREFIX + "any.json");
String content = new String(Files.readAllBytes(file.toPath()));

For java after 1.7
List<String> lines = Files.readAllLines(Paths.get(getClass().getResource("test.csv").toURI()));
Alternatively you can use Spring utils if you are in Spring echosystem
final val file = ResourceUtils.getFile("classpath:json/abcd.json");
To get to more behind the scene, check out following blog
https://todzhang.com/blogs/tech/en/save_resources_to_files

I faced the same issue.
The file was not found by a class loader, which means it was not packed into the artifact (jar). You need to build the project. For example, with maven:
mvn clean package
So the files you added to resources folder will get into maven build and become available to the application.
I would like to keep my answer: it does not explain how to read a file (other answers do explain that), it answers why InputStream or resource was null. Similar answer is here.

ClassLoader loader = Thread.currentThread().getContextClassLoader();
InputStream is = loader.getResourceAsStream("test.csv");
If you use context ClassLoader to find a resource then definitely it will cost application performance.

Now I am illustrating the source code for reading a font from maven created resources directory,
scr/main/resources/calibril.ttf
Font getCalibriLightFont(int fontSize){
Font font = null;
try{
URL fontURL = OneMethod.class.getResource("/calibril.ttf");
InputStream fontStream = fontURL.openStream();
font = new Font(Font.createFont(Font.TRUETYPE_FONT, fontStream).getFamily(), Font.PLAIN, fontSize);
fontStream.close();
}catch(IOException | FontFormatException ief){
font = new Font("Arial", Font.PLAIN, fontSize);
ief.printStackTrace();
}
return font;
}
It worked for me and hope that the entire source code will also help you, Enjoy!

Import the following:
import java.io.IOException;
import java.io.FileNotFoundException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;
import java.util.ArrayList;
The following method returns a file in an ArrayList of Strings:
public ArrayList<String> loadFile(String filename){
ArrayList<String> lines = new ArrayList<String>();
try{
ClassLoader classloader = Thread.currentThread().getContextClassLoader();
InputStream inputStream = classloader.getResourceAsStream(filename);
InputStreamReader streamReader = new InputStreamReader(inputStream, StandardCharsets.UTF_8);
BufferedReader reader = new BufferedReader(streamReader);
for (String line; (line = reader.readLine()) != null;) {
lines.add(line);
}
}catch(FileNotFoundException fnfe){
// process errors
}catch(IOException ioe){
// process errors
}
return lines;
}

getResource() was working fine with the resources files placed in src/main/resources only. To get a file which is at the path other than src/main/resources say src/test/java you need to create it exlicitly.
the following example may help you
import java.io.BufferedReader;
import java.io.FileReader;
import java.io.IOException;
import java.net.URISyntaxException;
import java.net.URL;
public class Main {
public static void main(String[] args) throws URISyntaxException, IOException {
URL location = Main.class.getProtectionDomain().getCodeSource().getLocation();
BufferedReader br = new BufferedReader(new FileReader(location.getPath().toString().replace("/target/classes/", "/src/test/java/youfilename.txt")));
}
}

You can use the com.google.common.io.Resources.getResource to read the url of file and then get the file content using java.nio.file.Files to read the content of file.
URL urlPath = Resources.getResource("src/main/resource");
List<String> multilineContent= Files.readAllLines(Paths.get(urlPath.toURI()));

if you are loading file in static method then
ClassLoader classLoader = getClass().getClassLoader();
this might give you an error.
You can try this
e.g. file you want to load from resources is resources >> Images >> Test.gif
import org.springframework.core.io.ClassPathResource;
import org.springframework.core.io.Resource;
Resource resource = new ClassPathResource("Images/Test.gif");
File file = resource.getFile();

To read the files from src/resources folder then try this :
DataSource fds = new FileDataSource(getFileHandle("images/sample.jpeg"));
public static File getFileHandle(String fileName){
return new File(YourClassName.class.getClassLoader().getResource(fileName).getFile());
}
in case of non static reference:
return new File(getClass().getClassLoader().getResource(fileName).getFile());

Does the code work when not running the Maven-build jar, for example when running from your IDE? If so, make sure the file is actually included in the jar. The resources folder should be included in the pom file, in <build><resources>.

The following class can be used to load a resource from the classpath and also receive a fitting error message in case there's a problem with the given filePath.
import java.io.InputStream;
import java.nio.file.NoSuchFileException;
public class ResourceLoader
{
private String filePath;
public ResourceLoader(String filePath)
{
this.filePath = filePath;
if(filePath.startsWith("/"))
{
throw new IllegalArgumentException("Relative paths may not have a leading slash!");
}
}
public InputStream getResource() throws NoSuchFileException
{
ClassLoader classLoader = this.getClass().getClassLoader();
InputStream inputStream = classLoader.getResourceAsStream(filePath);
if(inputStream == null)
{
throw new NoSuchFileException("Resource file not found. Note that the current directory is the source folder!");
}
return inputStream;
}
}

this.getClass().getClassLoader().getResource("filename").getPath()

My file in the test folder could not be found even though I followed the answers. It got resolved by rebuilding the project. It seems IntelliJ did not recognize the new file automatically. Pretty nasty to find out.

I got it work on both running jar and in IDE by writing as
InputStream schemaStream =
ProductUtil.class.getClassLoader().getResourceAsStream(jsonSchemaPath);
byte[] buffer = new byte[schemaStream.available()];
schemaStream.read(buffer);
File tempFile = File.createTempFile("com/package/schema/testSchema", "json");
tempFile.deleteOnExit();
FileOutputStream out = new FileOutputStream(tempFile);
out.write(buffer);

This worked pretty fine for me :
InputStream in = getClass().getResourceAsStream("/main/resources/xxx.xxx");
InputStreamReader streamReader = new InputStreamReader(in, StandardCharsets.UTF_8);
BufferedReader reader = new BufferedReader(streamReader);
String content = "";
for (String line; (line = reader.readLine()) != null;) {
content += line;
}

I get it to work without any reference to "class" or "ClassLoader".
Let's say we have three scenarios with the location of the file 'example.file' and your working directory (where your app executes) is home/mydocuments/program/projects/myapp:
a)A sub folder descendant to the working directory:
myapp/res/files/example.file
b)A sub folder not descendant to the working directory:
projects/files/example.file
b2)Another sub folder not descendant to the working directory:
program/files/example.file
c)A root folder:
home/mydocuments/files/example.file (Linux; in Windows replace home/ with C:)
1) Get the right path:
a)String path = "res/files/example.file";
b)String path = "../projects/files/example.file"
b2)String path = "../../program/files/example.file"
c)String path = "/home/mydocuments/files/example.file"
Basically, if it is a root folder, start the path name with a leading slash.
If it is a sub folder, no slash must be before the path name. If the sub folder is not descendant to the working directory you have to cd to it using "../". This tells the system to go up one folder.
2) Create a File object by passing the right path:
File file = new File(path);
3) You are now good to go:
BufferedReader br = new BufferedReader(new FileReader(file));

How can i write to a file when distributable file is launched?

I have program that read from a file. Now I created another task, that aims to update/write this file into same files. My problem now is when I generate my project's distributable file, during running and try to update my file, it does not write/update my changes. If I run it directly on my IDE, it works fine. This is wha I did so far:
private void tbleAddressMouseClicked(java.awt.event.MouseEvent evt) {
if(tbleAddress.getSelectedColumn()==3){
AddressUtil util = new AddressUtil();
List<AddressUtil> lists = util.getAddresses();
Address address = lists.get(tbleAddress.getSelectedRow());
if(tbleAddress.getValueAt(tbleAddress.getSelectedRow(), 0)!=null){
address.setRegion("\""+tbleAddress.getValueAt(tbleAddress.getSelectedRow(), 0).toString()+"\"");
}
lists.set(tbleAddress.getSelectedRow(), address);
try {
FileWriter fw;
fw = new FileWriter("./src/address.csv"); // This is where I doubt, if my **jar** file reads this directory
for(Address a:lists){
fw.append(a.getRegion);
fw.append(",");
fw.append(a.getAddressName());
fw.append("\n");
fw.flush();
}
fw.close();
} ...
My getAddresses is defined as:
public List<Address> getAddresses() {
List<Address> addresses = new ArrayList<>();
BufferedReader br = null;
InputStream in = this.getClass().getResourceAsStream("/address.csv");
BufferedReader reader = null;
try {
reader = new BufferedReader(new InputStreamReader(in));
String line = "";
while((line = reader.readLine())!=null){
String[] result = line.split(",");
Address address = new Address();
address.setRegion(result[0]);
address.setAddressName(result[1]);
addresses.add(address);
}...
My address.csv is of form:
"Region I","Sample St., Sample Address"
"Region II","Sample St., Sample Address 2"
...
Any help is much appreciated

Your issue with writing to the file is most likely due to the relative path you're using when you define the FileWriter. The dot in your path means "current working directory" so if your program is located at path C:\myProgram and you run your program such that it uses this path as the working directory then it's going to look for C:\myProgram\src\address.csv
So depending on the requirements for your program a relative path might be appropriate, in which case you need to determine what the correct path will be and ensure the file exists at that location, or you may to use some other mechanism to find the file.
I also notice that you're using getResourceAsStream to get an InputStream to your file. You should know that this only works if the file is available on the classpath of your program.

FileNotFoundException java cannot basic find file while it's there

I'm trying to read a basic txt file that contains prices in euros. My program is supposed to loop through these prices and then create a new file with the other prices. Now, the problem is that java says it cannot find the first file.
It is in the exact same package like this:
Java already fails at the following code:
FileReader fr = new FileReader("prices_usd.txt");
Whole code :
import java.io.*;
public class DollarToEur {
public static void main(String[] arg) throws IOException, FileNotFoundException {
FileReader fr = new FileReader("prices_usd.txt");
BufferedReader br = new BufferedReader(fr);
FileWriter fw = new FileWriter("prices_eur");
PrintWriter pw = new PrintWriter(fw);
String regel = br.readLine();
while(regel != null) {
String[] values = regel.split(" : ");
String beschrijving = values[0];
String prijsString = values[1];
double prijs = Double.parseDouble(prijsString);
double newPrijs = prijs * 0.913;
pw.println(beschrijving + " : " + newPrijs);
regel = br.readLine();
}
pw.close();
br.close();
}
}

Your file looks to be named "prices_usd" and your code is looking for "prices_usd.txt"

There are a couple of things you need to do:
Put the file directly under the project folder in Eclipse. When your execute your code in Eclipse, the project folder is considered to be the working directory. So you need to put the file there so that Java can find it.
Rename the file correctly with the .txt extn. From your screen print it looks like the file does not have an extension or may be it's just not visible.
Hope this helps!

It is bad practice to put resource files (like prices_usd.txt) in a package. Please put it under the resources/ directory. If you put it directly in the resources/ directory, you can access the file like this:
new FileReader(new File(this.getClass().getClassLoader().getResource("prices_usd.txt").getFile()));
But if you really have a good reason to put it in the package, you can access it like this:
new FileReader("src/main/java/week5/practicum13/prices_usd.txt");
But this will not work when you export your project (for example: as a jar).
EDIT 0: Also of course, your file's name needs to be "prices_usd.txt" and not just "prices_usd".
EDIT 1: The first (recommended) solution does return a string on .getFile() which can not directly be passed to the new File(...) constructor when the application is built / not run in the IDE. Spring has a solution to it though: org.springframework.core.io.ClassPathResource.
Simply use this code with Spring:
new FileReader(new ClassPathResource("prices_usd.txt").getFile());

downloading a file from src/main/resource folder [duplicate]

My project has the following structure:
/src/main/java/
/src/main/resources/
/src/test/java/
/src/test/resources/
I have a file in /src/test/resources/test.csv and I want to load the file from a unit test in /src/test/java/MyTest.java
I have this code which didn't work. It complains "No such file or directory".
BufferedReader br = new BufferedReader (new FileReader(test.csv))
I also tried this
InputStream is = (InputStream) MyTest.class.getResourcesAsStream(test.csv))
This also doesn't work. It returns null. I am using Maven to build my project.

Try the next:
ClassLoader classloader = Thread.currentThread().getContextClassLoader();
InputStream is = classloader.getResourceAsStream("test.csv");
If the above doesn't work, various projects have been added the following class: ClassLoaderUtil1 (code here).2
Here are some examples of how that class is used:
src\main\java\com\company\test\YourCallingClass.java
src\main\java\com\opensymphony\xwork2\util\ClassLoaderUtil.java
src\main\resources\test.csv
// java.net.URL
URL url = ClassLoaderUtil.getResource("test.csv", YourCallingClass.class);
Path path = Paths.get(url.toURI());
List<String> lines = Files.readAllLines(path, StandardCharsets.UTF_8);
// java.io.InputStream
InputStream inputStream = ClassLoaderUtil.getResourceAsStream("test.csv", YourCallingClass.class);
InputStreamReader streamReader = new InputStreamReader(inputStream, StandardCharsets.UTF_8);
BufferedReader reader = new BufferedReader(streamReader);
for (String line; (line = reader.readLine()) != null;) {
// Process line
}
Notes
See it in The Wayback Machine.
Also in GitHub.

Try:
InputStream is = MyTest.class.getResourceAsStream("/test.csv");
IIRC getResourceAsStream() by default is relative to the class's package.
As #Terran noted, don't forget to add the / at the starting of the filename

Try following codes on Spring project
ClassPathResource resource = new ClassPathResource("fileName");
InputStream inputStream = resource.getInputStream();
Or on non spring project
ClassLoader classLoader = getClass().getClassLoader();
File file = new File(classLoader.getResource("fileName").getFile());
InputStream inputStream = new FileInputStream(file);

Here is one quick solution with the use of Guava:
import com.google.common.base.Charsets;
import com.google.common.io.Resources;
public String readResource(final String fileName, Charset charset) throws IOException {
return Resources.toString(Resources.getResource(fileName), charset);
}
Usage:
String fixture = this.readResource("filename.txt", Charsets.UTF_8)

Non spring project:
String filePath = Objects.requireNonNull(getClass().getClassLoader().getResource("any.json")).getPath();
Stream<String> lines = Files.lines(Paths.get(filePath));
Or
String filePath = Objects.requireNonNull(getClass().getClassLoader().getResource("any.json")).getPath();
InputStream in = new FileInputStream(filePath);
For spring projects, you can also use one line code to get any file under resources folder:
File file = ResourceUtils.getFile(ResourceUtils.CLASSPATH_URL_PREFIX + "any.json");
String content = new String(Files.readAllBytes(file.toPath()));

For java after 1.7
List<String> lines = Files.readAllLines(Paths.get(getClass().getResource("test.csv").toURI()));
Alternatively you can use Spring utils if you are in Spring echosystem
final val file = ResourceUtils.getFile("classpath:json/abcd.json");
To get to more behind the scene, check out following blog
https://todzhang.com/blogs/tech/en/save_resources_to_files

I faced the same issue.
The file was not found by a class loader, which means it was not packed into the artifact (jar). You need to build the project. For example, with maven:
mvn clean package
So the files you added to resources folder will get into maven build and become available to the application.
I would like to keep my answer: it does not explain how to read a file (other answers do explain that), it answers why InputStream or resource was null. Similar answer is here.

ClassLoader loader = Thread.currentThread().getContextClassLoader();
InputStream is = loader.getResourceAsStream("test.csv");
If you use context ClassLoader to find a resource then definitely it will cost application performance.

Now I am illustrating the source code for reading a font from maven created resources directory,
scr/main/resources/calibril.ttf
Font getCalibriLightFont(int fontSize){
Font font = null;
try{
URL fontURL = OneMethod.class.getResource("/calibril.ttf");
InputStream fontStream = fontURL.openStream();
font = new Font(Font.createFont(Font.TRUETYPE_FONT, fontStream).getFamily(), Font.PLAIN, fontSize);
fontStream.close();
}catch(IOException | FontFormatException ief){
font = new Font("Arial", Font.PLAIN, fontSize);
ief.printStackTrace();
}
return font;
}
It worked for me and hope that the entire source code will also help you, Enjoy!

Import the following:
import java.io.IOException;
import java.io.FileNotFoundException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;
import java.util.ArrayList;
The following method returns a file in an ArrayList of Strings:
public ArrayList<String> loadFile(String filename){
ArrayList<String> lines = new ArrayList<String>();
try{
ClassLoader classloader = Thread.currentThread().getContextClassLoader();
InputStream inputStream = classloader.getResourceAsStream(filename);
InputStreamReader streamReader = new InputStreamReader(inputStream, StandardCharsets.UTF_8);
BufferedReader reader = new BufferedReader(streamReader);
for (String line; (line = reader.readLine()) != null;) {
lines.add(line);
}
}catch(FileNotFoundException fnfe){
// process errors
}catch(IOException ioe){
// process errors
}
return lines;
}

getResource() was working fine with the resources files placed in src/main/resources only. To get a file which is at the path other than src/main/resources say src/test/java you need to create it exlicitly.
the following example may help you
import java.io.BufferedReader;
import java.io.FileReader;
import java.io.IOException;
import java.net.URISyntaxException;
import java.net.URL;
public class Main {
public static void main(String[] args) throws URISyntaxException, IOException {
URL location = Main.class.getProtectionDomain().getCodeSource().getLocation();
BufferedReader br = new BufferedReader(new FileReader(location.getPath().toString().replace("/target/classes/", "/src/test/java/youfilename.txt")));
}
}

You can use the com.google.common.io.Resources.getResource to read the url of file and then get the file content using java.nio.file.Files to read the content of file.
URL urlPath = Resources.getResource("src/main/resource");
List<String> multilineContent= Files.readAllLines(Paths.get(urlPath.toURI()));

if you are loading file in static method then
ClassLoader classLoader = getClass().getClassLoader();
this might give you an error.
You can try this
e.g. file you want to load from resources is resources >> Images >> Test.gif
import org.springframework.core.io.ClassPathResource;
import org.springframework.core.io.Resource;
Resource resource = new ClassPathResource("Images/Test.gif");
File file = resource.getFile();

To read the files from src/resources folder then try this :
DataSource fds = new FileDataSource(getFileHandle("images/sample.jpeg"));
public static File getFileHandle(String fileName){
return new File(YourClassName.class.getClassLoader().getResource(fileName).getFile());
}
in case of non static reference:
return new File(getClass().getClassLoader().getResource(fileName).getFile());

Does the code work when not running the Maven-build jar, for example when running from your IDE? If so, make sure the file is actually included in the jar. The resources folder should be included in the pom file, in <build><resources>.

The following class can be used to load a resource from the classpath and also receive a fitting error message in case there's a problem with the given filePath.
import java.io.InputStream;
import java.nio.file.NoSuchFileException;
public class ResourceLoader
{
private String filePath;
public ResourceLoader(String filePath)
{
this.filePath = filePath;
if(filePath.startsWith("/"))
{
throw new IllegalArgumentException("Relative paths may not have a leading slash!");
}
}
public InputStream getResource() throws NoSuchFileException
{
ClassLoader classLoader = this.getClass().getClassLoader();
InputStream inputStream = classLoader.getResourceAsStream(filePath);
if(inputStream == null)
{
throw new NoSuchFileException("Resource file not found. Note that the current directory is the source folder!");
}
return inputStream;
}
}

this.getClass().getClassLoader().getResource("filename").getPath()

My file in the test folder could not be found even though I followed the answers. It got resolved by rebuilding the project. It seems IntelliJ did not recognize the new file automatically. Pretty nasty to find out.

I got it work on both running jar and in IDE by writing as
InputStream schemaStream =
ProductUtil.class.getClassLoader().getResourceAsStream(jsonSchemaPath);
byte[] buffer = new byte[schemaStream.available()];
schemaStream.read(buffer);
File tempFile = File.createTempFile("com/package/schema/testSchema", "json");
tempFile.deleteOnExit();
FileOutputStream out = new FileOutputStream(tempFile);
out.write(buffer);

This worked pretty fine for me :
InputStream in = getClass().getResourceAsStream("/main/resources/xxx.xxx");
InputStreamReader streamReader = new InputStreamReader(in, StandardCharsets.UTF_8);
BufferedReader reader = new BufferedReader(streamReader);
String content = "";
for (String line; (line = reader.readLine()) != null;) {
content += line;
}

I get it to work without any reference to "class" or "ClassLoader".
Let's say we have three scenarios with the location of the file 'example.file' and your working directory (where your app executes) is home/mydocuments/program/projects/myapp:
a)A sub folder descendant to the working directory:
myapp/res/files/example.file
b)A sub folder not descendant to the working directory:
projects/files/example.file
b2)Another sub folder not descendant to the working directory:
program/files/example.file
c)A root folder:
home/mydocuments/files/example.file (Linux; in Windows replace home/ with C:)
1) Get the right path:
a)String path = "res/files/example.file";
b)String path = "../projects/files/example.file"
b2)String path = "../../program/files/example.file"
c)String path = "/home/mydocuments/files/example.file"
Basically, if it is a root folder, start the path name with a leading slash.
If it is a sub folder, no slash must be before the path name. If the sub folder is not descendant to the working directory you have to cd to it using "../". This tells the system to go up one folder.
2) Create a File object by passing the right path:
File file = new File(path);
3) You are now good to go:
BufferedReader br = new BufferedReader(new FileReader(file));

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