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NoSuchFileException while running jar file

When we convert code to jar file we get this error.
the code works with IDE
public String getwordleString() {
Path path = Paths.get("..\\termproject\\word_database.txt");
List<String> wordList = new ArrayList<>();
try {
wordList = Files.readAllLines(path);
} catch (IOException e) {
e.printStackTrace();
}
Random random = new Random();
int position = random.nextInt(wordList.size());
return wordList.get(position).trim().toUpperCase();
}
Error part is : https://ibb.co/z6vZLW5

Using Paths.get("..\\termproject\\word_database.txt") assumes that there is a directory "..\termproject\word_database.txt" starting at the current working directory. If the file does not exist, you will get an error.
If you want to wrap the directory with the JAR, you can set you "termproject" file as a src or class file. Then this will be accessible via getClass().getResourceAsStream("/word_database.txt"). Now you can read the text from the file with a BufferedReader like so:
BufferedReader reader = new BufferedReader(new InputStreamReader(getClass().getResourceAsStream("/word_database.txt")));
String data = reader.lines().collect(Collectors.joining("\n"));

Read temporary file with FileReader and BufferedReader - Eclipse Java Plugin

I have a strange problem with read of a temporary file.
I create a new temp file:
File fileTemp = File.createTempFile("cex_data", ".txt");
After that, I write into it and if I call fileTemp.getAbsolutePath() and find it on disk is all ok. The file is created and written.
I call now another function:
readFromCexFile(fileTemp);
And in this function I would read file:
FileReader f;
BufferedReader b;
String filePath = fileTemp.getAbsolutePath();
//filePath = "C:\\tempPath/cex_data7121025199294655326.txt";
f = new FileReader(filePath);
b = new BufferedReader(f);
String s1;
while(true)
{
s1 = b.readLine();
....
if(s1=="")
break;
}
The problem was that is I use fileTemp.getAbsolutePath() doesn't read anything. However, if I use filePath = "C:\\tempPath/cex_data7121025199294655326.txt"; is all ok.
I tried also to print fileTemp.getAbsolutePath() and then replace "\" and "/" to be equals filePath = "C:\\tempPath/cex_data7121025199294655326.txt" syntax, but it doesn't work either.

Why don't you use this constructor.
f = new FileReader(fileTemp);

How to read from a file not in Eclipse in Java

I have a file which is needed for running tests - this file needs to be personalized (name and password) by whomever is running the test. I do not want to store this file in Eclipse (since it would need to be changed by whomever runs the test; also it would be storing personal info in the repo), so I have it in my home folder (/home/conrad/ssl.properties). How can I point my program to this file?
I've tried:
InputStream sslConfigStream = MyClass.class
.getClassLoader()
.getResourceAsStream("/home/" + name + "/ssl.properties");
I've also tried:
MyClass.class.getClassLoader();
InputStream sslConfigStream = ClassLoader
.getSystemResourceAsStream("/home/" + name + "/ssl.properties");
Both of these give me a RuntimeException because the sslConfigStream is null. Any help is appreciated!

Use a FileInputStream to read data from a file. The constructor takes a string path (or a File object, which encapsulates string path).
Note 1: A "resource" is a file which is in the classpath (alongside your java/class files). Since you don't want to store your file as a resource because you don't want it in your repo, ClassLoader.getSystemResourceAsStream() is not what you want.
Note 2: You should use a cross-platform way of getting a file in a home directory, as follows:
File homeDir = new File(System.getProperty("user.home"));
File propertiesFile = new File(homeDir, "ssl.properties");

InputStream sslConfigStream = new FileInputStream("/home/" + name + "/ssl.properties")

You can simplify your work, using Java's 7 method:
public static void main(String[] args) {
String fileName = "/path/to/your/file/ssl.properties";
try {
List<String> lines = Files.readAllLines(Paths.get(fileName),
Charset.defaultCharset());
for (String line : lines) {
System.out.println(line);
}
} catch (IOException e) {
e.printStackTrace();
}
}
You can also improve your way of reading properties file, using Properties class and forget about reading and parsing your .properties file:
http://www.mkyong.com/java/java-properties-file-examples/

Is this a graphics program (ie. using the Swing library)? If so it is a pretty simple task of using a JFileChooser.
http://docs.oracle.com/javase/6/docs/api/javax/swing/JFileChooser.html
JFileChooser f = new JFileChooser();
int rval = f.showOpenDialog(this);
if (rval == JFileChooser.APPROVE_OPTION) {
// Do something with file called f
}

You can also use Scanner to read the file.
String fileContent = "";
try {
Scanner scan = new Scanner(
new File( System.getProperty("user.home")+"/ssl.properties" ));
while(scan.hasNextLine()) {
fileContent += scan.nextLine();
}
scan.close();
} catch(FileNotFoundException e) {
}

Android: Open file with specific path [duplicate]

I have a filename in my code as :
String NAME_OF_FILE="//sdcard//imageq.png";
FileInputStream fis =this.openFileInput(NAME_OF_FILE); // 2nd line
I get an error on 2nd line :
05-11 16:49:06.355: ERROR/AndroidRuntime(4570): Caused by: java.lang.IllegalArgumentException: File //sdcard//imageq.png contains a path separator
I tried this format also:
String NAME_OF_FILE="/sdcard/imageq.png";

The solution is:
FileInputStream fis = new FileInputStream (new File(NAME_OF_FILE)); // 2nd line
The openFileInput method doesn't accept path separators.
Don't forget to
fis.close();
at the end.

This method opens a file in the private data area of the application. You cannot open any files in subdirectories in this area or from entirely other areas using this method. So use the constructor of the FileInputStream directly to pass the path with a directory in it.

openFileInput() doesn't accept paths, only a file name
if you want to access a path, use File file = new File(path) and corresponding FileInputStream

I got the above error message while trying to access a file from Internal Storage using openFileInput("/Dir/data.txt") method with subdirectory Dir.
You cannot access sub-directories using the above method.
Try something like:
FileInputStream fIS = new FileInputStream (new File("/Dir/data.txt"));

You cannot use path with directory separators directly, but you will
have to make a file object for every directory.
NOTE: This code makes directories, yours may not need that...
File file= context.getFilesDir();
file.mkdir();
String[] array=filePath.split("/");
for(int t=0; t< array.length -1 ;t++)
{
file=new File(file,array[t]);
file.mkdir();
}
File f=new File(file,array[array.length-1]);
RandomAccessFileOutputStream rvalue = new RandomAccessFileOutputStream(f,append);

String all = "";
try {
BufferedReader br = new BufferedReader(new FileReader(filePath));
String strLine;
while ((strLine = br.readLine()) != null){
all = all + strLine;
}
} catch (IOException e) {
Log.e("notes_err", e.getLocalizedMessage());
}

File file = context.getFilesDir();
file.mkdir();
String[] array = filePath.split("/");
for(int t = 0; t < array.length - 1; t++) {
file = new File(file, array[t]);
file.mkdir();
}
File f = new File(file,array[array.length- 1]);
RandomAccessFileOutputStream rvalue =
new RandomAccessFileOutputStream(f, append);

I solved this type of error by making a directory in the onCreate event, then accessing the directory by creating a new file object in a method that needs to do something such as save or retrieve a file in that directory, hope this helps!
public class MyClass {
private String state;
public File myFilename;
#Override
protected void onCreate(Bundle savedInstanceState) {//create your directory the user will be able to find
super.onCreate(savedInstanceState);
if (Environment.MEDIA_MOUNTED.equals(state)) {
myFilename = new File(Environment.getExternalStorageDirectory().toString() + "/My Directory");
if (!myFilename.exists()) {
myFilename.mkdirs();
}
}
}
public void myMethod {
File fileTo = new File(myFilename.toString() + "/myPic.png");
// use fileTo object to save your file in your new directory that was created in the onCreate method
}
}

I did like this
var dir = File(app.filesDir, directoryName)
if(!dir.exists()){
currentCompanyFolder.mkdir()
}
var directory = app.getDir(directoryName, Context.MODE_PRIVATE)
val file = File(directory, fileName)
file.outputStream().use {
it.write(body.bytes())
}

Java, reading a file from current directory?

I want a java program that reads a user specified filename from the current directory (the same directory where the .class file is run).
In other words, if the user specifies the file name to be "myFile.txt", and that file is already in the current directory:
reader = new BufferedReader(new FileReader("myFile.txt"));
does not work. Why?
I'm running it in windows.

Try
System.getProperty("user.dir")
It returns the current working directory.

The current directory is not (necessarily) the directory the .class file is in. It's working directory of the process. (ie: the directory you were in when you started the JVM)
You can load files from the same directory* as the .class file with getResourceAsStream(). That'll give you an InputStream which you can convert to a Reader with InputStreamReader.
*Note that this "directory" may actually be a jar file, depending on where the class was loaded from.

None of the above answer works for me. Here is what works for me.
Let's say your class name is Foo.java, to access to the myFile.txt in the same folder as Foo.java, use this code:
URL path = Foo.class.getResource("myFile.txt");
File f = new File(path.getFile());
reader = new BufferedReader(new FileReader(f));

Files in your project are available to you relative to your src folder. if you know which package or folder myfile.txt will be in, say it is in
----src
--------package1
------------myfile.txt
------------Prog.java
you can specify its path as "src/package1/myfile.txt" from Prog.java

If you know your file will live where your classes are, that directory will be on your classpath. In that case, you can be sure that this solution will solve your problem:
URL path = ClassLoader.getSystemResource("myFile.txt");
if(path==null) {
//The file was not found, insert error handling here
}
File f = new File(path.toURI());
reader = new BufferedReader(new FileReader(f));

Thanks #Laurence Gonsalves your answer helped me a lot.
your current directory will working directory of proccess so you have to give full path start from your src directory like mentioned below:
public class Run {
public static void main(String[] args) {
File inputFile = new File("./src/main/java/input.txt");
try {
Scanner reader = new Scanner(inputFile);
while (reader.hasNextLine()) {
String data = reader.nextLine();
System.out.println(data);
}
reader.close();
} catch (FileNotFoundException e) {
System.out.println("scanner error");
e.printStackTrace();
}
}
}
While my input.txt file is in same directory.

Try this:
BufferedReader br = new BufferedReader(new FileReader("java_module_name/src/file_name.txt"));

try using "."
E.g.
File currentDirectory = new File(".");
This worked for me