Data d=DateFormat.parse(String s)
SimpleDateFormat f=new SimpleDataFormat(String geshi)
geshi如下:
"yyyy.MM.dd G 'at' hh:mm:ss z" ->> 1996.07.10 AD at 15:08:56 PDT
"EEE, MMM d, ''yy" ->> Wed, July 10, '96
"h:mm a" ->> 12:08 PM
"hh 'o''clock' a, zzzz" ->> 12 o'clock PM, Pacific Daylight Time
"K:mm a, z" ->> 0:00 PM, PST
"yyyyy.MMMMM.dd GGG hh:mm aaa" ->> 1996.July.10 AD 12:08 PM
之后调用
f.format(d)
SimpleDateFormat f=new SimpleDataFormat(String geshi)
geshi如下:
"yyyy.MM.dd G 'at' hh:mm:ss z" ->> 1996.07.10 AD at 15:08:56 PDT
"EEE, MMM d, ''yy" ->> Wed, July 10, '96
"h:mm a" ->> 12:08 PM
"hh 'o''clock' a, zzzz" ->> 12 o'clock PM, Pacific Daylight Time
"K:mm a, z" ->> 0:00 PM, PST
"yyyyy.MMMMM.dd GGG hh:mm aaa" ->> 1996.July.10 AD 12:08 PM
之后调用
f.format(d)
Date date1 = sdf1.parse("99/11/23");SimpleDateFormat sdf2 = new SimpleDateFormat("MM dd,yy");
Date date2 = sdf2.parse("July 10,96");
Date date1 = sdf1.parse("99/11/23");SimpleDateFormat sdf2 = new SimpleDateFormat("MM dd,yy");
Date date2 = sdf2.parse("July 10,96");
你的代码有问题:non-static method parse(java.lang.String) cannot be referenced from a static context
unreported exception: java.text.ParseException; must be caught or declared to be thrown
class SimpleDataFormat not found in class ShowDates
你的代码有问题:
能给出正确的吗?
import java.util.*;
import java.text.*;public class test12
{
public static void main(String args[])
{
DateFormat df=DateFormat.getInstance();
try
{
Date d=df.parse("2002-10-10");
}
catch(Exception e)
{
}
}
}
妈的,javaDoc不准嘛,上面的OK了
import java.util.*;
import java.text.*;public class test12
{
public static void main(String args[])
{
DateFormat df=DateFormat.getInstance();
try
{
Date d=df.parse("2002-10-10");
}
catch(Exception e)
{
}
}
}
妈的,javaDoc不准嘛,上面的OK了
import java.util.*;//the method :
public String getTime(){
Calendar myCalendar=Calendar.getInstance();
String yyyy=String.valueOf(myCalendar.get(Calendar.YEAR));
String mm=String.valueOf(myCalendar.get(Calendar.MONTH)+1);
String dd=String.valueOf(myCalendar.get(Calendar.DAY_OF_MONTH));
String hh=String.valueOf(myCalendar.get(Calendar.HOUR_OF_DAY));
String ff=String.valueOf(myCalendar.get(Calendar.MINUTE));
String ss=String.valueOf(myCalendar.get(Calendar.SECOND));
mm=(1==mm.length())?("0"+mm):mm;
dd=(1==dd.length())?("0"+dd):dd;
hh=(1==hh.length())?("0"+hh):hh;
ff=(1==ff.length())?("0"+ff):ff;
ss=(1==ss.length())?("0"+ss):ss;
return yyyy+"--"+mm+"--"+dd+"--"+hh+"--"+ff+"--"+ss;
}
给分!!
在重复一编:
例如: 用户输入:99/11/23 转化为 1999-11-23 13:24:30
用户输入:July 10,96 转化为 1996-07-10
用户输入:2002.03.24 转化为 2003-03-24 是不是应该根据用户的输入先解析为统一的格式再进行格式转换?
Date 数据类型,然后利用SimpleDateFormat转化成其它的格式
{
public static void main(String args[])
{
DateFormat df=DateFormat.getInstance();
try
{
Date d=df.parse("2002-10-10");
}
catch(Exception e)
{
}
}
}
完成String到Date的转化,属于类型转化
然后利用SimpleDateFormat进行格式转化啊
DateFormat df=DateFormat.getInstance();
try
{
Date d=df.parse("2002-10-10");
}
catch(Exception e)
{
}
SimpleDateFormat sdf=new SimpleDateFormat("yyyy-MM-dd");
String ss=sdf.format(d);
System.out.println(ss );
出错信息:
java.lang.NullPointerException at java.util.Calendar.setTime(Calendar.java:882) at java.text.SimpleDateFormat.format
(SimpleDateFormat.java:400) at java.text.DateFormat.format(DateFormat.java:305) at ShowDates.main(ShowDates.java:74)Exception in thread "main"
Date d=df.parse("2002-10-10");该了很多次
难道字符串不能转化成日期格式吗?我在看看,请高手指教。
public class test12
{
public static void main(String args[])
{
SimpleDateFormat sdf1 = new SimpleDateFormat("yyyy.MM.dd hh:mm:ss");
SimpleDateFormat sdf2 = new SimpleDateFormat("yyyy-mm-dd");
DateFormat df=DateFormat.getInstance();
Date d=null;
Date test=new Date();
try
{
String s=" 2002.11.11 11:11:11 ";
//d=df.parse("1996.July.10 AD 12:08 PM");
String str=sdf1.format(test);
System.out.println("first"+sdf1.format(test));
//d=df.parse("July 10,96");
//System.out.println("second"+sdf2.format(d));
d=df.parse(str);
System.out.println("third"+sdf2.format(d));
}
catch(Exception e)
{
System.out.println(e.getMessage());
}
}
}
真上奇怪了,难道parse函数不能使用了吗?
Date d=new Date()
d.setYear(int )
d.setMonth(int)
d.setDate(int)
d.setHours(int)
d.setMinutes(int)
d.setSeconds(int)
当然要首先解析用户输入的字符串,得到年,月等。
用下面两个函数,你不会让我给你编写吧
int index=s.indexOf(int ch, int fromIndex)
s.substring(int beginIndex, int endIndex)
比如用户输入 : 92,01.13 如何解析?
比如用户输入: July,12,2002 又如何解析?
public class test13
{
public static void main(String args[])
{
String s="2222-12-22";
String str=null;
int i,j;
int year;
int month;
int day;
Character a=new Character('-');
int ch=a.hashCode();
i=0;
int index=s.indexOf(ch, i);
if (index!=-1)
{
str=s.substring(i, index);
if (str.length()!=4)System.out.println("格式不正确");
else
{
year=Integer.parseInt(str);
s=s.substring(index+1,s.length());
index=s.indexOf(ch,i);
}
}
else
{
System.out.println("格式不正确");
}
}
}