我有一个表示日期的字符串。
String date_s = "2011-01-18 00:00:00.0";
我想把它转换成一个日期，并以YYYY-MM-DD格式输出。
2011-01-18
我怎样才能做到这一点呢?
好吧，根据我在下面找到的答案，以下是我尝试过的一些方法:
String date_s = " 2011-01-18 00:00:00.0";
SimpleDateFormat dt = new SimpleDateFormat("yyyyy-mm-dd hh:mm:ss");
Date date = dt.parse(date_s);
SimpleDateFormat dt1 = new SimpleDateFormat("yyyyy-mm-dd");
System.out.println(dt1.format(date));
但是它输出02011-00-1而不是所需的2011-01-18。我做错了什么?
当前回答
try
{
String date_s = "2011-01-18 00:00:00.0";
SimpleDateFormat simpledateformat = new SimpleDateFormat("yyyy-MM-dd HH:mm:ss.S");
Date tempDate=simpledateformat.parse(date_s);
SimpleDateFormat outputDateFormat = new SimpleDateFormat("yyyy-MM-dd");
System.out.println("Output date is = "+outputDateFormat.format(tempDate));
} catch (ParseException ex)
{
System.out.println("Parse Exception");
}
其他回答
格式是大小写敏感的，所以使用MM表示月，而不是MM(这是分钟)和yyyy 作为参考，您可以使用以下备忘单。
G Era designator Text AD
y Year Year 1996; 96
Y Week year Year 2009; 09
M Month in year Month July; Jul; 07
w Week in year Number 27
W Week in month Number 2
D Day in year Number 189
d Day in month Number 10
F Day of week in month Number 2
E Day name in week Text Tuesday; Tue
u Day number of week (1 = Monday, ..., 7 = Sunday) Number 1
a Am/pm marker Text PM
H Hour in day (0-23) Number 0
k Hour in day (1-24) Number 24
K Hour in am/pm (0-11) Number 0
h Hour in am/pm (1-12) Number 12
m Minute in hour Number 30
s Second in minute Number 55
S Millisecond Number 978
z Time zone General time zone Pacific Standard Time; PST; GMT-08:00
Z Time zone RFC 822 time zone -0800
X Time zone ISO 8601 time zone -08; -0800; -08:00
例子:
"yyyy.MM.dd G 'at' HH:mm:ss z" 2001.07.04 AD at 12:08:56 PDT
"EEE, MMM d, ''yy" Wed, Jul 4, '01
"h:mm a" 12:08 PM
"hh 'o''clock' a, zzzz" 12 o'clock PM, Pacific Daylight Time
"K:mm a, z" 0:08 PM, PDT
"yyyyy.MMMMM.dd GGG hh:mm aaa" 02001.July.04 AD 12:08 PM
"EEE, d MMM yyyy HH:mm:ss Z" Wed, 4 Jul 2001 12:08:56 -0700
"yyMMddHHmmssZ" 010704120856-0700
"yyyy-MM-dd'T'HH:mm:ss.SSS'Z'" 2001-07-04T12:08:56.235-0700
"yyyy-MM-dd'T'HH:mm:ss.SSSXXX" 2001-07-04T12:08:56.235-07:00
"YYYY-'W'ww-u" 2001-W27-3
你也可以使用substring()
String date_s = "2011-01-18 00:00:00.0";
date_s.substring(0,10);
如果你想在日期前留出空格，那就用吧
String date_s = " 2011-01-18 00:00:00.0";
date_s.substring(1,11);
String str = "2000-12-12";
Date dt = null;
SimpleDateFormat formatter = new SimpleDateFormat("yyyy-MM-dd");
try
{
dt = formatter.parse(str);
}
catch (Exception e)
{
}
JOptionPane.showMessageDialog(null, formatter.format(dt));
请参阅此处“日期及时间模式”。http://docs.oracle.com/javase/7/docs/api/java/text/SimpleDateFormat.html
import java.text.SimpleDateFormat;
import java.util.Date;
import java.text.ParseException;
public class DateConversionExample{
public static void main(String arg[]){
try{
SimpleDateFormat sourceDateFormat = new SimpleDateFormat("yyyy-MM-DD HH:mm:ss");
Date date = sourceDateFormat.parse("2011-01-18 00:00:00.0");
SimpleDateFormat targetDateFormat = new SimpleDateFormat("yyyy-MM-dd");
System.out.println(targetDateFormat.format(date));
}catch(ParseException e){
e.printStackTrace();
}
}
}
从格式中删除一个y:
SimpleDateFormat dt1 = new SimpleDateFormat("yyyyy-mm-dd");
它应该是:
SimpleDateFormat dt1 = new SimpleDateFormat("yyyy-mm-dd");
