Develop Reference

Java : how to create a directory inside under usr in linux from a java app?

I'm trying to create a folder under /usr in linux from a java program. Here's what I've done. I understand that I lack the permission to do so under /usr but what do I need to add?
public void createDirectory (String path)
{
File directory = new File(path);
if (!directory.exists()) {
if (!directory.mkdirs()) {
System.out.println("couldn't create file");
}
}
}
Here the sysout statement gets printed. what needs to be done here? Would greatly appreciate your help and thanks in advance.

mkdirs() is used in case you want to create nested folders.
Try with mkdir() instead:
public void createDirectory (String path)
{
File directory = new File(path);
if (!directory.exists()) {
if (!directory.mkdir()) {
System.out.println("couldn't create file");
}
}
}
Please note that you must provide the full path in order to make it work. Also as #Reimeus mentioned in the comment above, is not a good idea to write or create anything at that level, I would suggest to create it under /home/your_user/

Inputting a text file into a program?

I had a lecture today on inputting and outputting but it didn't really seem to explain where the text file is etc..
here is my code:
package inputoutput;
import java.util.*;
import java.io.*;
public class input {
public static void main(String[] args) throws FileNotFoundException {
String name;
int lineCount = 0;
File input = new File("lab1task3.txt");
Scanner in = new Scanner(input);
while(in.hasNextLine()){
lineCount++;
}
System.out.println(lineCount);
}
}
I get a file not found exception but the text file is in the same folder as the program?

Please first read up on the difference between relative and absolute paths. An absolute path is:
C:\Users\Ceri\workspace1\inputoutput\src\inputoutput\lab1task3.txt
A relative path would be just "lab1task3.txt", which is what is given. That means that lab1task3.txt can be found relative to the working directory (e.g if the working directory was "C:\Users\Ceri\workspace1\inputoutput\src\inputoutput\" then it would find it).
However, you could also use an absolute path, but remember that doing so means that it will only work if a file is in the same place on the machine running it. E.g, if you submit with "C:\Users\Ceri\workspace1\inputoutput\src\inputoutput\" in your code then it will only work if someone else has that same file and location on their computer. Please note that if this is an assignment, the module convenor/marker probably does not have afolder called C:\Users\Ceri.... If you submit your work using a relative path, anyone using your code just needs to make sure the file is relatively in the same place (e.g in the same folder).
If this doesn't matter, you need to escape the back slash characters with another back slash in the path. This should work:
package inputoutput;
import java.util.*;
import java.io.*;
public class input {
public static void main(String[] args) throws FileNotFoundException {
String name;
int lineCount = 0;
File input = new File("C:\\Users\\Ceri\\workspace1\\inputoutput\\src\\inputoutput\\lab1task3.txt");
Scanner in = new Scanner(input);
while(in.hasNextLine()){
lineCount++;
}
System.out.println(lineCount);
}
}
I notice you are using eclipse. Your "working directory" is your workspace. Therefore you want to move your file to:
C:\Users\Ceri\workspace1\inputoutput\lab1task3.txt
This should work for you using a "relative" path which you had in your opening post.

You're confusing class file location and the "user's working directory", the latter being what Java uses to determine the root of the file path (unless absolute paths are needed), and you can find its location easily via:
System.out.println(System.getProperty("user.dir"));
I advise you to forgo use of files altogether when all you need to do is read in data, and instead get the text file as a program resource:
// where you swap the name of your class for MyClass
InputStream fileResource = MyClass.class.getResourceAsStream("myFile.txt");
Scanner scanner = new Scanner(inputStream);
Note that if you must use a File, then find out what the user's working directory is, as shown above, and then tailor your file path so that it is relative to this working directory.
Try:
File file = new File("src/inputoutput/lab1task3.txt");

My guess is that your current working directory is not the same place as the project location. If your working directory were, the file would definitely be found if it does indeed have that name.
To workaround this issue you can always be using a InputStream instead, like so:
InputStream inputStream = new InputStream("lab1task3.txt");
Scanner scanner = new Scanner(inputStream);
If you want to see your current working directory you can use something like this:
public class JavaApplication1 {
public static void main(String[] args) {
System.out.println("Working Directory = " +
System.getProperty("user.dir"));
}
}

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;
}`

Using path to load the file

this one thing just keeps coming back to me. I am trying to load the .csv file in Java and it looks like this:
public static List<String> getEntryList () throws IOException{
final String NAME = "test.csv";
final String PATH = "resources/csvFiles";
final Path path = FileSystems.getDefault().getPath(PATH, NAME);
return Files.readAllLines(path, Charset.forName("UTF-8"));
}
obviously not working. Gives me java.nio.file.NoSuchFileException: resources\csvFiles\test.csv exception, but works ok when I enter tehe full path on the drive. I tried using classpath but also didn't work. What is the proper way of doing this?

Try this:
InputStream is = YourClassName.class.getResourceAsStream("test.csv");

Text input and output java

I am trying to read 2 files after i read the files i want to get their contents and manipulate the contents of the two files then update a new file which is the output. The files are in the same folder as the program but the program always throws a FileNotFoundException.
Below is my code:-
import java.io.*;
import java.util.Scanner;
public class UpdateMaster {
public static void main(String[] args)
{
String master = "Customer.dat";
String trans = "Transactns.dat";
String newMaster = "Temp.txt";
Scanner inputStreamMaster = null;
Scanner inputStreamTrans = null;
PrintWriter inputStreamNewMaster = null;
try
{
inputStreamMaster = new Scanner(new File(master));
inputStreamTrans = new Scanner(new File(trans));
inputStreamNewMaster = new PrintWriter(newMaster);
}
catch(FileNotFoundException e)
{
System.out.println("Error: you opend a file that does not exist.");
System.exit(0);
}
catch(IOException e)
{
System.out.println("Error.");
System.exit(0);
}
do
{
String transLine = inputStreamTrans.nextLine();
String masterLine = inputStreamMaster.nextLine();
String[] transLineArr = transLine.split(",");
String[] masterLineArr = masterLine.split(",");
int trAccNo = Integer.parseInt(transLineArr[0]);
int sales = Integer.parseInt(transLineArr[1]);
int masterAccNo = Integer.parseInt(masterLineArr[0]);
int balance = Integer.parseInt(masterLineArr[1]);
while(masterAccNo== trAccNo){
inputStreamNewMaster.println(trAccNo+ " , "+masterAccNo);
masterLine = inputStreamMaster.nextLine();
masterLineArr = masterLine.split(",");
masterAccNo = Integer.parseInt(masterLineArr[0]);
balance = Integer.parseInt(masterLineArr[1]);
}
balance = balance + sales;
inputStreamNewMaster.println(masterAccNo+ " , "+balance);
}while(inputStreamTrans.hasNextLine());
inputStreamMaster.close();
inputStreamTrans.close();
inputStreamNewMaster.close();
//System.out.println(" the line were written to "+ newMaster);
}
}

Like #Ankit Rustagi said in the comments, you need the full path of the files if you want to keep the current implementation.
However, there is a solution where you only need the file names: use BufferedReader / BufferedWriter. See here an example on how to use these classes (in the example it uses the full path but it works without it too).

Use absolute path
String master = "C:/Data/Customer.dat";
String trans = "C:/Data/Transactns.dat";
String newMaster = "C:/Data/Temp.txt";

The code works for me, i guess you misspelled some filename(s) or your files are in the wrong folder. I created your files on the same level as the src or the project. Also this is the folder where the files are exspected.

There's nothing wrong with using relative paths like tihis. What's happening is that your program is looking for the files in the directory where you execute the program, which doesn't have to be the folder of the program. You can confirm this by logging the absolute path of the files before you try to read them. For example:
File masterFile = new File(master);
System.out.printf("Using master file '%s'%n", masterFile.getAbsolutePath());
inputStreamMaster = new Scanner(masterFile);
In general you should not hardcode file paths but allow the user to specify them in someway, for example using command line arguments, a configuration file with a well known path, or an interactive user interface.
There is a way to locate the program's class file but it's a little tricky because Java allows classes to be loaded from compressed archives that may be located in remote systems. It's better to solve this problem in some other manner.

Try this:
String current = new java.io.File( "." ).getCanonicalPath();
System.out.println("I look for files in:"+current);
To see what directory your program expects to find its input files. If it shows the correct directory, check spelling of filenames. Otherwise, you have a clue as to what's gone wrong.