java - java.util.日历

Question

我的应用程序：

System.out.print("Please enter date (and time): ");
myTask.setWhen(
   input.nextInt(),
   input.nextInt(),
   input.nextInt(),
   input.nextInt(),
   input.nextInt());

我的二传手：

public void setWhen(int year, int month, int date, int hourOfDay, int minute){
    this.year = year;
    this.month = month;
    this.date = date;
    this.hourOfDay = hourOfDay;
    this.minute = minute;

但是，当它准备好让用户输入日期和时间时，它会引发异常。另外，如果用户输入 4/7/2013 1:30pm 而不是 4、7、2013、13、30，会发生什么？谢谢。

Answer 1

代码应该是这样的（只是快速的想法）：

System.out.print("Please enter date (and time): ");
String inputText;
Scanner scanner = new Scanner(System.in);
inputText = scanner.nextLine();
scanner.close();
//parse input and parsed put as input parameter for your setwhen() method
setWhen(myOwnParserForInputFromConsole(inputText));

Answer 2

使用标准的另一种方法：

import java.text.DateFormat;
import java.text.ParseException;
import java.util.Calendar;
import java.util.Date;

public class Calend {

   public static void main( String[] args ) throws ParseException {
      Calendar cal = Calendar.getInstance();
      DateFormat formatter =
         DateFormat.getDateTimeInstance(
            DateFormat.FULL, DateFormat.FULL );
      DateFormat scanner;
      Date date;

      scanner = DateFormat.getDateTimeInstance(
         DateFormat.SHORT, DateFormat.SHORT );
      date = scanner.parse( "7/4/2013 21:01:05" );
      cal.setTime( date );
      System.out.println( formatter.format( cal.getTime()));

      scanner = DateFormat.getDateTimeInstance(
         DateFormat.MEDIUM, DateFormat.MEDIUM );
      date = scanner.parse( "7 avr. 2013 21:01:05" );
      cal.setTime( date );
      System.out.println( formatter.format( cal.getTime()));

      scanner = DateFormat.getDateTimeInstance(
         DateFormat.FULL, DateFormat.FULL );
      date = scanner.parse( "dimanche 7 avril 2013 21 h 01 CEST" );
      cal.setTime( date );
      System.out.println( scanner.format( cal.getTime()));
   }
}

如果你想使用不同的语言环境，它在 DateFormat 中存在一个构造函数来处理它。

Answer 3

从javadoc中，nextInt抛出“ InputMismatchException- 如果下一个标记与 Integer 正则表达式不匹配，或者超出范围”。这意味着您不能盲目调用nextInt并希望输入是int.

您可能应该将输入读取为 aString并对该输入执行检查，以查看它是否有效。

public static void main(String[] args) throws IOException {
    final Scanner scanner = new Scanner(System.in);
    //read year
    final String yearString = scanner.next();
    final int year;
    try {
        year = Integer.parseInt(yearString);
        //example check, pick appropriate bounds
        if(year < 2000 || year > 3000) throw new NumberFormatException("Year not in valid range");
    } catch (NumberFormatException ex) {
        throw new RuntimeException("Failed to parse year.", ex);
    }
    final String monthString = scanner.next();
    final int month;
    try {
        month = Integer.parseInt(monthString);
        //example check, pick appropriate bounds
        if(month < 1 || month > 12) throw new NumberFormatException("Month not in valid range");
    } catch (NumberFormatException ex) {
        throw new RuntimeException("Failed to parse month.", ex);
    }
    //and the rest of the values
}

然后，当您拥有所有输入并且已知它们有效时，请调用setWhen.

显然，您可以尝试再次读取该数字，而不是抛出异常。