Spinner的getSelectedItemPosition()方法返回意外结果

Question

我的应用程序有一个微调器,里面有四个条目.我需要能够检索选择了哪些标签,而不是实际标签本身.以下是我执行此操作的部分代码:

// Set up the activity's Spinner
    spinnerAdapter = ArrayAdapter.createFromResource(this, R.array.pay_periods, android.R.layout.simple_spinner_item);
    spinnerAdapter.setDropDownViewResource(android.R.layout.simple_spinner_dropdown_item);
    Spinner s = (Spinner) findViewById(R.id.main_spinner_payperiod);
    s.setAdapter(spinnerAdapter);

在以后的方法中:

switch(payPeriod.getSelectedItemPosition()){
    case(0): // Daily
        dailyAllowance = Float.parseFloat(payment.getText().toString());
    case(1): // Weekly
        dailyAllowance = (Float.parseFloat(payment.getText().toString())) / 7;
    case(2): // Bi-weekly
        dailyAllowance = (Float.parseFloat(payment.getText().toString())) / 14;
    case(3): // 30 days
        dailyAllowance = (Float.parseFloat(payment.getText().toString())) / 30;
    case(Spinner.INVALID_POSITION):
        dailyAllowance = 0;
    default:
        dailyAllowance = 42;    // Junk value, for debugging purposes
    }

无论我选择哪个Spinner项目,这个方法似乎总是返回42.任何人都可以帮我找出原因吗？谢谢!

Answer 1

把break你的switch-case条件

switch(payPeriod.getSelectedItemPosition()) { 

    case(0): // Daily
        dailyAllowance = Float.parseFloat(payment.getText().toString());
    break;

    case(1): // Weekly
        dailyAllowance = (Float.parseFloat(payment.getText().toString())) / 7;
    break;

    case(2): // Bi-weekly
        dailyAllowance = (Float.parseFloat(payment.getText().toString())) / 14;
    break;

    case(3): // 30 days
        dailyAllowance = (Float.parseFloat(payment.getText().toString())) / 30;
    break;

    case(Spinner.INVALID_POSITION):
        dailyAllowance = 0;
    break;
    default:
        dailyAllowance = 42;    // Junk value, for debugging purposes
    break;
}

break在交换机情况下省略可能会产生错误的行为.如果选择case(不break),则流程仍然继续,并且default始终执行条件.