Java SimpleDateFormat给出了错误的日期
我正在输入一个特定的日期(dd/MM/yyyy),在显示该日期时,输出的是其他内容。Java SimpleDateFormat给出了错误的日期,java,date,simpledateformat,Java,Date,Simpledateformat,我正在输入一个特定的日期(dd/MM/yyyy),在显示该日期时,输出的是其他内容。 import java.text.*; import java.io.*; import java.util.*; class InvalidUsernameException extends Exception //Class InvalidUsernameException { InvalidUsernameException(String s) { super(
import java.text.*;
import java.io.*;
import java.util.*;
class InvalidUsernameException extends Exception //Class InvalidUsernameException
{
InvalidUsernameException(String s)
{
super(s);
}
}
///////////////////////////////////////////////////////////////////////////////////////////
class InvalidPasswordException extends Exception //Class InvalidPasswordException
{
InvalidPasswordException(String s)
{
super(s);
}
}
///////////////////////////////////////////////////////////////////////////////////////////
class InvalidDateException extends Exception //Class InvalidPasswordException
{
InvalidDateException(String s)
{
super(s);
}
}
///////////////////////////////////////////////////////////////////////////////////////////
class EmailIdb1 //Class Email Id b1
{
String username, password;
int domainid;
Date dt;
EmailIdb1()
{
username = "";
domainid = 0;
password = "";
dt = new Date();
}
EmailIdb1(String u, String pwd, int did, int d, int m, int y)
{
username = u;
domainid = did;
password = pwd;
dt = new Date(y,m,d); // I think There is a problem
SimpleDateFormat formater = new SimpleDateFormat ("yyyy/MM/dd"); //Or there can be a problem
try{
if((username.equals("User")))
{
throw new InvalidUsernameException("Invalid Username");
}
else if((password.equals("123")))
{
throw new InvalidPasswordException("Invalid Password");
}
else{
System.out.println("\nSuccesfully Login on Date : "+formater.format(dt));
}
}
catch(Exception e)
{
}
}
}
///////////////////////////////////////////////////////////////////////////////////////////
class EmailId //Class Email Id
{
public static void main(String args[])
{
int d,m,y,did;
String usn,pwd;
EmailIdb1 eml;
try{
usn = args[0];
pwd = args[1];
did = Integer.parseInt(args[2]);
d = Integer.parseInt(args[3]);
m = Integer.parseInt(args[4]);
y = Integer.parseInt(args[5]);
switch(m)
{
case 2: if(d==29 && y%4 == 0)
{
eml = new EmailIdb1(usn,pwd,did,d,m,y);
}
else if(d<=28 && d>=1)
{
eml = new EmailIdb1(usn,pwd,did,d,m,y);
}
else{
throw new InvalidDateException("Wrong Date.");
}
break;
case 1: case 3: case 5: case 7: case 8: case 10:
case 12: if(d>=1 && d<=31)
{
eml = new EmailIdb1(usn,pwd,did,d,m,y);
}
else
{
throw new InvalidDateException("Invalid Date");
}
break;
case 4: case 6: case 9:
case 11: if(d>=1 && d<=30)
{
eml = new EmailIdb1(usn,pwd,did,d,m,y);
}
else
{
throw new InvalidDateException("Invalid Date");
}
break;
default : throw new InvalidDateException("Invalid Date");
}
}
catch(InvalidDateException ed)
{
System.out.println(ed);
}
}
}
因为输入是
Successfully Login on Date : 3894/06/04
首先
new Date(int year, int month, int date)
已弃用-您不应该使用它
其次,根据javadoc:
/**
* Allocates a <code>Date</code> object and initializes it so that
* it represents midnight, local time, at the beginning of the day
* specified by the <code>year</code>, <code>month</code>, and
* <code>date</code> arguments.
*
* @param year the year minus 1900.
* @param month the month between 0-11.
* @param date the day of the month between 1-31.
* @see java.util.Calendar
* @deprecated As of JDK version 1.1,
* replaced by <code>Calendar.set(year + 1900, month, date)</code>
* or <code>GregorianCalendar(year + 1900, month, date)</code>.
*/
因此,如果你通过1994年,你将得到日期为“3894年”。如果你想得到“1994”,你应该通过94作为一年。
月份表示为0-11范围内的整数,因此如果超过5,则在本例中格式为“06”,因为5表示六月而不是五月。您不应将日期构造函数与三个整数一起使用,因为它不受欢迎。
SimpleDateFormat
您如何处理?此外,永远不要使用空的挡块。另外,对代码的注释充其量也无济于事,充其量也会损害可读性。阅读文档,int,int):年份-年份减去1900。@Boristeider Yeah man..这些注释也相当令人不安-太多的代码。这是你的责任剥离你的代码,使一个。谢谢,一个真正的帮助。。。
/**
* Allocates a <code>Date</code> object and initializes it so that
* it represents midnight, local time, at the beginning of the day
* specified by the <code>year</code>, <code>month</code>, and
* <code>date</code> arguments.
*
* @param year the year minus 1900.
* @param month the month between 0-11.
* @param date the day of the month between 1-31.
* @see java.util.Calendar
* @deprecated As of JDK version 1.1,
* replaced by <code>Calendar.set(year + 1900, month, date)</code>
* or <code>GregorianCalendar(year + 1900, month, date)</code>.
*/