在这种情况下，使用多个else if语句是不好的做法吗？

Question

我正在使用多个else ifs来使程序根据微调器中当前选择的值来执行某些操作，但是我不禁想到这样做的一种更简洁的方法。 有什么建议么？

    if(spinnerinput.equals(spinnerinput2)) {
        output.setText(input.getText());
    }

    else if(spinnerinput.equals("Base 2") && spinnerinput2.equals("Base 10")) {
        String regex = "[0-1]+";

        if (input.getText().toString().matches(regex)) {
            output.setText(binaryToDecimal(input.getText().toString()));
        } else {
            Context context = getApplicationContext();
            CharSequence text = "Invalid characters for a binary number!";
            int duration = Toast.LENGTH_SHORT;

            Toast toast = Toast.makeText(context, text, duration);
            toast.show();
        }
    }

    else if(spinnerinput.equals("Base 10") && spinnerinput2.equals("Base 2")) {
        String regex = "[0-9]+";

        if(input.getText().toString().matches(regex)) {
            output.setText(decimalToBinary(input.getText().toString()));
        } else {
            Context context = getApplicationContext();
            CharSequence text = "Invalid characters for a decimal number!";
            int duration = Toast.LENGTH_SHORT;

            Toast toast = Toast.makeText(context, text, duration);
            toast.show();
        }
    }

Answer 1

您的if-else问题不多，但是您可以将代码重构到最少。 您底部的两个条件块的区别仅在于CharSequence文本。 为什么不将其转换为方法并再次使用相同的代码？

private void yourCurrentMethod() {

    if(spinnerinput.equals(spinnerinput2)) {
        output.setText(input.getText());
    }

    else if(spinnerinput.equals("Base 2") && spinnerinput2.equals("Base 10")) {
        String regex = "[0-1]+";
        boolean isBinary = true;
        doSomething(isBinary, "Invalid characters for a binary number!");
    }

    else if(spinnerinput.equals("Base 10") && spinnerinput2.equals("Base 2")) {
        String regex = "[0-9]+";
        boolean isBinary = false;
        doSomething(isBinary, "Invalid characters for a decimal number!");
    }
}

private void doSomething(boolean isBinary, CharSequence text){
    if(input.getText().toString().matches(regex)) {
        if(isBinary){
            output.setText(binaryToDecimal(input.getText().toString()));
        } else {
            output.setText(decimalToBinary(input.getText().toString()));
        } 
    } else {
        Context context = getApplicationContext();
        int duration = Toast.LENGTH_SHORT;

        Toast toast = Toast.makeText(context, text, duration);
        toast.show();
    }
}