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
public class SystemDateTest {
String stringDate;
public static void main(String[] args) {
SystemDateTest systemDateTest = new SystemDateTest();
// format date into String
SimpleDateFormat simpleDateFormat = new SimpleDateFormat("dd-MM-yyyy hh:mm:ss");
systemDateTest.setStringDate(simpleDateFormat.format(systemDateTest.getDate()));
System.out.println(systemDateTest.getStringDate());
}
public Date getDate() {
return new Date();
}
public String getStringDate() {
return stringDate;
}
public void setStringDate(String stringDate) {
this.stringDate = stringDate;
}
}
其他回答
public class SystemDateTest {
String stringDate;
public static void main(String[] args) {
SystemDateTest systemDateTest = new SystemDateTest();
// format date into String
SimpleDateFormat simpleDateFormat = new SimpleDateFormat("dd-MM-yyyy hh:mm:ss");
systemDateTest.setStringDate(simpleDateFormat.format(systemDateTest.getDate()));
System.out.println(systemDateTest.getStringDate());
}
public Date getDate() {
return new Date();
}
public String getStringDate() {
return stringDate;
}
public void setStringDate(String stringDate) {
this.stringDate = stringDate;
}
}
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
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));
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");
为什么不简单地使用它呢
Date convertToDate(String receivedDate) throws ParseException{
SimpleDateFormat formatter = new SimpleDateFormat("dd-MM-yyyy");
Date date = formatter.parse(receivedDate);
return date;
}
还有，这是另一种方式:
DateFormat df = new SimpleDateFormat("dd/MM/yyyy");
String requiredDate = df.format(new Date()).toString();
or
Date requiredDate = df.format(new Date());
