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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());

Accessing a file in a runnable JAR [duplicate]

This question already has answers here:
Java resource as File
(6 answers)
Closed 9 years ago.
What I am attempting to do is store a text file (that won't change) inside the JAR of the program so that it can be read. The purpose of the text file is that it will be read in by one of my classes and the contents of the text file will be added to an JEditorPane. The file will basically be a tutorial and when the user clicks on the option to read the tutorial, the file contents will be read and displayed in a new window that pops up.
I have the GUI portion of it down, but as far as storing the file in the JAR so it can be accessed, I am at a lost. I've read that using an InputStream will work, but after trying a few things I haven't gotten it to work yet.
I also store images in the JAR to be used as icons for the GUI windows. This is accomplished with:
private Image icon = new ImageIcon(getClass()
.getResource("resources/cricket.jpg")).getImage();
But, this doesn't work when trying to get a file:
private File file = new File(getClass.getResource("resources/howto.txt"));
Here is my Class as it is now:
public class HowToScreen extends JFrame{
/**
*
*/
private static final long serialVersionUID = -3760362453964229085L;
private JEditorPane howtoScreen = new JEditorPane("text/html", "");
private Image icon = new ImageIcon(getClass().getResource("resources/cricket.jpg")).getImage();
private BufferedReader txtReader = new BufferedReader(new InputStreamReader(getClass().getResourceAsStream("/resources/howto.txt")));
public HowToScreen(){
setSize(400,300);
setLocation(500,200);
setTitle("Daily Text Tutorial");
setIconImage(icon);
howtoScreen.setEditable(false);
howtoScreen.setText(importFileStream());
add(howtoScreen);
setVisible(true);
}
public String importFile(){
String text = "";
File file = new File("howto.txt");
Scanner in = null;
try {
in = new Scanner(file);
} catch (FileNotFoundException e) {
e.printStackTrace();
}
while(in.hasNext()){
text += in.nextLine();
}
in.close();
return text;
}
public String importFileStream(){
String text = "";
Scanner in = new Scanner(txtReader);
while(in.hasNext()){
text += in.nextLine();
}
in.close();
return text;
}
}
Ignore the importFile method as that is being removed in favor of storing the tutorial file inside the JAR, making the program wholly self contained as I am limited to how much space the program can use.
EDIT:
After trying all of the suggestions below, I checked to see if my JAR is packaging the text file in it and it is not. When opening the JAR with 7zip, in my resources folder the picture I use for icons is there, but not the text file.

You cannot use File inside a JAR file. You need to use InputStream to read the text data.
BufferedReader txtReader = new BufferedReader(new InputStreamReader(getClass().getResourceAsStream("/resources/mytextfile.txt")));
// ... Use the buffered reader to read the text file.

Try the next (with the full path package):
InputStream inputStream = ClassLoader.getSystemClassLoader().
getSystemResourceAsStream("com/company/resources/howto.txt");
InputStreamReader streamReader = new InputStreamReader(inputStream, "UTF-8");
BufferedReader in = new BufferedReader(streamReader);
for (String line; (line = in.readLine()) != null;) {
// do something with the line
}

You code will not compile. Class.getResource() returns a URL, and File has no constructor with a URL as an argument.
You can just use .getResourceAsStream() instead, it returns an InputStream directly, you just have to read the contents of the file from that stream.
Note: both of these methods return null if the resource is not found: don't forget to check for that...

the contents of the text file will be added to an JEditorPane.
See DocumentVewer & especially JEditorPane.setPage(URL).
Since the help is an embedded-resource it will be necessary to gain an URL using getResource(String) as detailed in the info. page.
.. tried this: URL url = this.getClass().getResource("resources/howto.txt");
Change:
URL url = this.getClass().getResource("resources/howto.txt");
To:
URL url = this.getClass().getResource("/resources/howto.txt"); // note leading '/'

java FileInputStream cannot find file

I'm very new at coding java and I'm having a lot of difficulty.
I'm suppose to write a program using bufferedreader that reads from a file, that I have already created named "scores.txt".
So I have a method named processFile that is suppose to set up the BufferedReader and loop through the file, reading each score. Then, I need to convert the score to an integer, add them up, and display the calculated mean.
I have no idea how to add them up and calculate the mean, but I'm currently working on reading from the file.
It keeps saying that it can't fine the file, but I know for sure that I have a file in my documents named "scores.txt".
This is what I have so far...it's pretty bad. I'm just not so good at this :( Maybe there's is a different problem?
public static void main(String[] args) throws IOException,
FileNotFoundException {
String file = "scores.txt";
processFile("scores.txt");
//calls method processFile
}
public static void processFile (String file)
throws IOException, FileNotFoundException{
String line;
//lines is declared as a string
BufferedReader inputReader =
new BufferedReader (new InputStreamReader
(new FileInputStream(file)));
while (( line = inputReader.readLine()) != null){
System.out.println(line);
}
inputReader.close();
}

There are two main options available
Use absolute path to file (begins from drive letter in Windows or
slash in *.nix). It is very convenient for "just for test" tasks.
Sample
Windows - D:/someFolder/scores.txt,
*.nix - /someFolder/scores.txt
Put file to project root directory, in such case it will be visible
to class loader.

Place the scores.txt in the root of your project folder, or put the full path to the file in String file.
The program won't know to check your My Documents folder for scores.txt

If you are using IntelliJ, create an input.txt file in your package and right click the input.txt file and click copy path. You can now use that path as an input parameter.
Example:
in = new FileInputStream("C:\\Users\\mda21185\\IdeaProjects\\TutorialsPointJava\\src\\com\\tutorialspoint\\java\\input.txt");

Take the absolute path from the local system if you'r in eclipse then right-click on the file and click on properties you will get the path copy it and put as below this worked for me In maven project keep the properties file in src/main/resources `
private static Properties properties = new Properties();
public Properties simpleload() {
String filepath="C:/Users/shashi_kailash/OneDrive/L3/JAVA/TZA/NewAccount/AccountConnector/AccountConnector-DEfgvf/src/main/resources/sample.properties";
try(FileInputStream fis = new FileInputStream(filepath);) {
//lastModi = propFl.lastModified();
properties.load(fis);
} catch (Exception e) {
System.out.println("Error loading the properties file : sample.properties");
e.printStackTrace();
}
return properties;
}`

Add docx file in resources and create executable jar

I want to add
docx files in resources folder, use those files in code written in class located at another package of same application.
And then I want to make an executable jar out of it which will be working on windows.
I read its not easy to make such jar :( and there is no fool prroof way...
I have tried searching for it on net and found I will have to create URL and then file and then use it...
however, when I use below code, I am not able to get URL itself...
URL urlOfDraftInSamePackage = CreateDraft.class.getResource("Draft_in_same_package.docx");
System.out.println("urlOfDraftInSamePackage is "+urlOfDraftInSamePackage.toString());
//This prints : urlOfDraftInSamePackage is file:/D:/aditya_workspace/SampleDraftMaker/bin/draftProcessing/Draft_in_same_package.docx
URL urlOfDraftInResourceFolder = CreateDraft.class.getResource("resouces/Draft_Apartment.docx");
System.out.println("urlOfDraftInResourceFolder is "+urlOfDraftInResourceFolder.toString());
//this gives null pointer exception
URI uri = null;
try {
uri = urlOfDraftInSamePackage.toURI();
File file = new File(uri);
System.out.println("file made");
} catch (URISyntaxException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
below is my folder structure:
can anyone pls help me in creating such executable jar using eclipse?
Thanks In Advance!!!

Following code works for me:
public static void testResource() throws IOException {
InputStream stream = Deserializace.class.getResourceAsStream("resources/ser.log");
BufferedReader reader = new BufferedReader(new InputStreamReader(stream));
String s;
while ( (s = reader.readLine()) != null) {
System.out.println(s);
}
}
Build directory structure:
Test.class
resources/ser.log
You must ensure that your resource directory is copied to correct place.

File Delete and Rename in Java

I have the following Java code which will search in an xml for a specific tag and then will add some text to it and save that file. I couldnt find a way to rename the emporary file to the original file. Please suggest.
import java.io.*;
class ModifyXML {
public void readMyFile(String inputLine) throws Exception
{
String record = "";
File outFile = new File("tempFile.tmp");
FileInputStream fis = new FileInputStream("InfectiousDisease.xml");
BufferedReader br = new BufferedReader(new InputStreamReader(fis));
FileOutputStream fos = new FileOutputStream(outFile);
PrintWriter out = new PrintWriter(fos);
while ( (record=br.readLine()) != null )
{
if(record.endsWith("<add-info>"))
{
out.println(" "+"<add-info>");
out.println(" "+inputLine);
}
else
{
out.println(record);
}
}
out.flush();
out.close();
br.close();
//Also we need to delete the original file
//outFile.renameTo(InfectiousDisease.xml);//Not working
}
public static void main (String[] args) {
try
{
ModifyXML f = new ModifyXML();
f.readMyFile("This is infectious disease data");
}
catch(Exception e)
{
e.printStackTrace();
}
}
}
Thanks

First delete the original file and then rename the new file:
File inputFile = new File("InfectiousDisease.xml");
File outFile = new File("tempFile.tmp");
if(inputFile.delete()){
outFile.renameTo(inputFile);
}

A good method to rename files is.
File file = new File("path-here");
file.renameTo(new File("new path here"));

In your code there are several issues.
First your description mentions renameing the original file and adding some text to it. Your code doesn't do that, it opens two files, one for reading and one for writing (with the additional text). That is the right way to do things, as adding text in-place is not really feasible using the techniques you are using.
The second issue is that you are opening a temporary file. Temporary files remove themselves upon closing, so all the work you did adding your text disappears as soon as you close the file.
The third issue is that you are modifying XML files as plain text. This sometimes works as XML files are a subset of plain text files, but there is no indication that you attempted to ensure that the output file was an XML file. Perhaps you know more about your input files than is mentioned, but if you want this to work correctly for 100% of the input cases, you probably want to create a SAX writer that writes out all a SAX reader reads, with the additional information in the correct tag location.

You can use
outFile.renameTo(new File(newFileName));
You have to ensure these files are not open at the time.