如何在每个步骤之后检查相同条件，而又不用一遍又一遍地重复Java中的相同代码？

Question

In my console application, I give the user the option to type "exit" at any point to return to the main menu. As the user enters data, i prompt him/her through the console for various things, and collect the data using scanner .

My question is how can I check to see if the user entered "exit" after each prompt (as opposed to the requested information) without having to use the same if statement after each step?

As I see it, any kind of while or for loops are insufficient because they only check the condition at the beginning, when I need to check the condition between inputs, and I need each input/prompt to execute only once per iteration.

The key here is that each prompt/code executed between the checks is DIFFERENT, so loops won't work.

Here is some example code:

String first;
String second;

Scanner input = new Scanner(System.in);


//prompt user for input
first = input.nextline();

if(first.equals("exit")){
    //return to start menu
    input.close();
    return;
}

//prompt user for DIFFERENT input
second = input.nextline()

if(second.equals("exit")){
    //return to start menu
    input.close();
    return;
}

Answer 1

Write a method...

public boolean isExit(String value) {
    return value.equals("exit");
}

You can then check this method each time...

String value = input.nextLine();
if (!isExit(value) {
    // Handle the normal text
} else {
    // Handle the exit operations...
}

You could put additional code in the check, but I would prefer to have an additional method that handles the exit operation...for example...

String value = input.nextLine();
if (!isExit(value) {
    // Handle the normal text
} else {
    doExit();
}

Take a look at Defining Methods for more details...

Updated

Focus on the idea that a method should do a single job and have no side effects...

Having said that, I would setup my code in such away that if the user enters exit at the prompt, the method can exit of it's own accord, without the need for return; statement...

For example...

public int selectYourMeal() {
    // Prompt...
    int option = -1;
    String value = input.nextLine();
    if (!isExit(value) {
        // Handle the normal text
    } else {
        option = EXIT_OPTION;
    }
    return option;  
}

Where EXIT_OPTION is a special value, which the caller and identify and deal with as it sees fit.

I'm also old school, in that I was taught that a method should have one entry and one exit point, you really want to avoid having multiple exit points within your methods, as it becomes very difficult to follow the logic...

Answer 2

If I understand you correctly, I recommend a do-while loop. It will first take text the first time, perform an action, and if it is exit it will break the loop, otherwise it will repeat.

 do{
 text = input.nextLine();

 //whatever code you want here to perform with input

 }while(!(text.equals("exit"));

Answer 3

I would suggest you use an List<String> of words. You can return it. Also, it's a really bad idea to close a Scanner wrapping System.in , once you do you cannot re-open it.

List<String> words = new ArrayList<>();
Scanner input = new Scanner(System.in);
for (int i = 0; input.hasNextLine(); i++) {
    String line = input.nextLine();
    words.add(line);
    if (line.equalsIgnoreCase("exit")) {
        break;
    }
}
System.out.println(words);
return words;