What is the location of file that is read?

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Im testing an example code which shows the use of exceptions management, and it uses a text file as example.

My problem is that i don't know where is the default location of that archive to create it.

This is the code:

package exceptionManager;

import java.io.FileInputStream;
import java.io.FileNotFoundException;
import java.io.IOException;
public class CheckedExceptionDemo {
    public static void main(String[] args) {
        //Below line calls readFile method and prints content of it
        String filename="test.txt";
        try {
            String fileContent = new CheckedExceptionDemo().readFile(filename);
            System.out.println(fileContent);
        } catch (FileNotFoundException e) {
            System.out.println("File:"+ filename+" is missing, Please check file name");
        } catch (IOException e) {
            System.out.println("File is not having permission to read, please check the permission");
        }
    }
    public String readFile(String filename)throws FileNotFoundException, IOException{
        FileInputStream fin;
        int i;
        String s="";
        fin = new FileInputStream(filename);
        // read characters until EOF is encountered
        do {
        i = fin.read();
        if(i != -1)  s =s+(char) i+"";
        } while(i != -1);
        fin.close();
        return s;
    }
}

When no absolute paths are defined, the paths are normally relative to the working directory - meaning if you run the application from /myfolder, it will be read from /myfolder/test.txt.

As such, it is helpful to use standards (like System.getProperty("user.home")) as prefix.

Note that defining an absolute path will most likely prevent the application from working System independent. The same goes for using / or \ as path segment separator.

Chapter 8 – Reading and Writing Files, The path specifies the location of a file on the computer. For example, there is a file on my Windows 7 laptop with the filename project.docx in the path C:\Users\� To edit this type of read-only file, you'd want to edit the file in a folder that does allow editing, and then move the newly created file into the original file's folder, overwriting the original. For example, a common location for read-only files is C:\Windows\System32\drivers\etc , which stores the hosts file .

You can check your currently user directory with:

    System.out.println(System.getProperty("user.dir"));

Python 3 Notes: File Path and CWD, On this page: open(), file path, CWD ('current working directory'), r 'raw string' prefix, os.getcwd(), os.chdir(). Below, you are opening up a file for reading: In Windows, the default location is often 'C:/program Files (x86)/Python35-32' ( which� When you access a file on an operating system, a file path is required. The file path is a string that represents the location of a file. It’s broken up into three major parts: Folder Path: the file folder location on the file system where subsequent folders are separated by a forward slash / (Unix) or backslash \ (Windows)

The file name actually is a relative path - it does not contain any drive letter and does not start with / or \. It is resolved starting from your current working directory. This directory depends on how you actually start the program.

E.g. if you run java from a command line prompt, then the directory you are currently in is the base directory. If you cd to another directory, the program will search for the file there.

If you start the program from your IDE, the default folder for the working directory is the project's base directory. You can change this in the run-configuration, though.

If nothing helps, try printing the directory from your program, e.g.:

System.out.println("Current dir: " + new java.io.File( "." ).getCanonicalPath())

Reading file from external shared location, I am trying to read file from a shared location. The filepath is a UNC- "\\server\ folder\file.xml". I need to read this in java code. Our Pega server is installed on linux� The computer file hosts is an operating system file that maps hostnames to IP addresses.It is a plain text file. Originally a file named HOSTS.TXT was manually maintained and made available via file sharing by Stanford Research Institute for the ARPANET membership, containing the hostnames and address of hosts as contributed for inclusion by member organizations.

Get the path of running file (.py) in Python: __file, In Python, you can get the location (path) of the running script file .py with __file__. __file__ is useful for reading other files based on the� One reason your file doesn't open in any of the ways described above is that you're not actually dealing with a PEM file. You might instead have a file that just uses a similarly spelled file extension. When that's the case, there isn't a necessity for the two files to be related or for them to work with the same software programs.

(Tutorial) Reading and Writing Files in Python, Learn how to read and write data into flat files, such as CSV, JSON, text The exact location of the record can be known using the index of that� The oratab file can be edited by the oracle user using vi or another text editor. Each line in the oratab file has three elements separated by colons. The first element is the SID, the second indicates the Oracle Home directory for that SID, and the third indicates if the database should be started and stopped by the dbstart/dbshut commands.

A path to a folder directory or a file name is a string of folder names where a particular file is located. For example the true path to my pictures is C:Documents and SettingsYour user nameMy DocumentsMy Pictures.

Comments
  • It is from the current working directory.
  • @Sambit can you clarify a little bit how did you get into this ? Thank you so much
  • The current working directory is not actually a Java concept; it is a feature of every operating system. If you were to enter type test.txt in a Windows terminal or cat test.txt in a Unix terminal, the same rules would apply.
  • Thanks, i got a copy of my "test.txt" file in every folder from that path you said (System.getProperty("user.home")) to below, but when i run my code it still throws FileNotFoundException. Could you please check if it works on your machine?
  • It does - substituting String filename = "text.txt"; with String filename = System.getProperty("user.home") + File.separator + "test.txt"; led to File contents being printed on the console.
  • I see clearly that your response is perfect, but idk why it still doesn't work for me :/
  • Try a simple System.out.println(filename) and see if a file at that location exists. It is very easy to introduce typos when typing manually as opposed to selecting from a menu. If all else fails, do the following: new FileWriter("myFilename.txt").write("I am a stegosaurus!"); and then locate the file with the name myFilename.txt
  • It's always safe to use a slash / in java as path segment separator. It works for both, Windows and Linux/Unix. Do not use `\` to separate the paths - it only works for Windows and you have to escape the backslash in a string. I do not recommend to make heavy use of system properties and File.separator, it just makes your code unreadable. It's OK to say "./files/test.txt". It just works.