Java: Reading from files; trying to create a program that will help me to organize assignments by subjects?

Question

When user enters serial number/subject id, the program reads from a bunch of files and finds the corresponding .txt file and displays the name of assignments listed in the .txt file. I'm having a lot of trouble creating the program please help me; this is the code I have come up with.
import java.io.IOException;
import java.util.Scanner;
import java.io.File;
public class prog{
public void prog(){
System.out.print("Enter a serial number.");
Scanner in;
try{
if (I.equals("English07")){
in = new Scanner(new File("English07"));
File f = new File(“C:\Documents and Settings\Administrator\Desktop.txt”);
String I = new String ("English07");
I = in.nextLine();
System.out.println(I);
}
if(II.equals("chemistry07")){
in = new Scanner(new File("chemistry07"));
File f = new File(“C:\Documents and Settings\Administrator\Desktop.txt”);
String II = new String ("chemistry07");
II = in.nextLine();
System.out.println(II);
}
}catch(IOException i){
System.out.println("Error: " + i.getMessage());
}
}
}

-TY

Answer 1

Some recommendations:

(1) Don't take user input in a constructor method. Constructors should be quick initialization, in general.

(2) Don't write identical code over and over; refactor it into a separate method, where it can be written and maintained in only one place.

(3) Don't hard-wire the same constant into the code (C:\Documents and Settings ...) multiple times.

(4) Don't play around with Administrator's files, if you don't have to. Use a folder in C or something like that.

(5) It's almost never necessary to write "String s = new String("foo");". In almost every case "String s = "foo";" will work just fine. And in your code, the "String l = new String()" is immediately overwritten with l = in.nextLine(), so the initialization is completely unnecessary. Java doesn't require you to pre-allocate space for Strings.

-------------------------------------------------
How about this:

import java.io.BufferedReader;
import java.io.File;
import java.io.FileInputStream;
import java.io.FileNotFoundException;
import java.io.IOException;
import java.io.InputStreamReader;

public class prog {

// This is the directory to search
static final String BASE_DIR = "C:\\prog\\";

// This is the method to scan a file and print its first line
void scan(String aFilename) throws FileNotFoundException, IOException {
File inFile = new File(BASE_DIR + aFilename + ".txt");
if (! inFile.exists()) {
System.out.println("File " + inFile + " does not exist.");
return;
}
BufferedReader in = new BufferedReader( new InputStreamReader( new FileInputStream( inFile)));
// Print out the first line only.
// Is that what you wanted?
// It's what your code does, though.
String s = null;
while ((s = in.readLine()) != null) System.out.println(s);
in.close();
}

// This takes a line of user input
// Return false if user hit end-of-file
boolean queryAndPrint() throws FileNotFoundException, IOException {
System.out.print("Enter serial number: ");
String s = (new BufferedReader(new InputStreamReader( System.in ) ) ).readLine();
if (s == null) return false;
scan(s);
return true;
}

// This is the main method which executes forever
public static void main(String[] args) {
prog p = new prog();
try {
while (p.queryAndPrint()) ;
} catch (Throwable th) {
th.printStackTrace();
}
}
}

Answer 2

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Answer 3

Try this

import java.io.*;

public class myprog
{
public static void main(String args[])
{
BufferedReader br,in;
String line;

System.out.println("Enter Serial Number");
try{
br = new BufferedReader(new InputStreamReader(System.in));
in = new BufferedReader(new FileReader(br.readLine() + ".txt"));
while((line = in.readLine()) != null)
{
System.out.println(line);
}
br.close();
in.close();
}catch (IOException e){
e.printStackTrace();
}
}
}

Run on test files in same and other directories.
Use another directory by adding path to beginning of serial number - user can control which directory to search in this way:

Enter Serial Number
c:\serial\new\1409080

and the contents of c:\serial\new\1409080.txt will be output.