Control Flow in Multiple Method Program

Question

I'm writing a simple Java program, to familiarise myself with methods, that logs the people a person has met. For instance, if I meet Alison in London, I would log the following (Format: name,gender,when,where):

Alison,F,apr2013,London

The program is built up as follows:

The user is presented with different opportunities:

1. Log person
2. Search for all people named [name]
3. Search for all people met [place]
4. Get a list of available commands
5. Quit

Here is the skeleton of my code:

    public void chooseCommand() throws FileNotFoundException {
    System.out.println("Enter command: ");
    text = input.next();
    myCommand = Integer.parseInt(text);

    while (myCommand !=5) {
        if (myCommand == 1) {
            writeToFile();    //Log new person
        }
        // Search for person type
        else if (myCommand == 2) {
            searchFor();    // Search for person by name
        }
        // Search for place
        else if (myCommand == 3) {
            searchFor();    // Search for person by place
        }
        // help
        else if (myCommand == 4) {
            showCommands();  // get a list of available commands        
        }
        else if (myCommand == 5) {
            exit();
        }
        // default
        else {
            System.out.println("Command not found");
        }
    }
}

This works just fine. However, after I choose one of the five options (log new person, search for name, search for place, help, quit), I would like to go back to the chooseCommand() method, so as to get the same options presented again, instead of having the initially chosen method loop infinitely. That is to say, after I log a new person, I want to be able to get new options, as opposed to having to log new people for all eternity, without killing the program.

    // REGISTER
public void writeToFile() {
    // Write to file
    try {
        BufferedWriter output = new BufferedWriter(new FileWriter(file, true)); 
        System.out.println("Enter sighting: ");

        for (int i = 0; i < personBlank.length; i++) {
            System.out.println(personInfo[i] + ": ");
            personEntry = input.next();
            personBlank[i] = personEntry;
        }

        // change to toString() method
        observed = personBlank[0] + "," + personBlank[1] + "," + personBlank[2] + "," + personBlank[3];

        if (observed.equals(escape)) {
            exit();
        }
        else {
            output.write(observed);    // log new person 
            output.newLine();
            output.close();
        }
        back();
    }
    catch (IOException e){
        e.printStackTrace();
    }
}

Any help on this is highly appreciated!

Answer 1

public void someMethod() {
  while(isRunning) {

  chooseCommand();

  }
}

Then in chooseCommand() lose the loop, make option 5 set isRunning = false instead of exit(), and use a switch statement for prettyness.

e.g.

public void chooseCommand() throws FileNotFoundException {
    System.out.println("Enter command: ");
    text = input.next();
    myCommand = Integer.parseInt(text);

    switch (myCommand) {
        case 1: 
            writeToFile();    //Log new person
            break;

        case 2:
        // Search for place
            break;

        case 3: 
            searchFor();    // Search for person by place
            break;

        // help
        case 4:
            showCommands();  // get a list of available commands 
            break;       

        case 5: 
            this.isRunning = false;
            break;

        default:
            System.out.println("Command not found");

    }
}

Answer 2

in the place of your code where chooseCommand() was called you could use a boolean and   check that boolean is true to call chooseCommand()

java pseudocode
------------------
boolean continue=true;

while(continue)
{
System.out.println("Do you want to continue?");

Scanner scan=new Scanner(System.in);

if(scan.nextLine().equals("true"))

chooseCommand();

else 

continue = false;
}