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
/**
* Method will take Date in "MMMM, dd yyyy HH:mm:s" format and return time difference like added: 3 min ago
*
* @param date : date in "MMMM, dd yyyy HH:mm:s" format
* @return : time difference
*/
private String getDurationTimeStamp(String date) {
String timeDifference = "";
//date formatter as per the coder need
SimpleDateFormat sdf = new SimpleDateFormat("MMMM, dd yyyy HH:mm:s");
TimeZone timeZone = TimeZone.getTimeZone("EST");
sdf.setTimeZone(timeZone);
Date startDate = null;
try {
startDate = sdf.parse(date);
} catch (ParseException e) {
MyLog.printStack(e);
}
//end date will be the current system time to calculate the lapse time difference
Date endDate = new Date();
//get the time difference in milliseconds
long duration = endDate.getTime() - startDate.getTime();
long diffInSeconds = TimeUnit.MILLISECONDS.toSeconds(duration);
long diffInMinutes = TimeUnit.MILLISECONDS.toMinutes(duration);
long diffInHours = TimeUnit.MILLISECONDS.toHours(duration);
long diffInDays = TimeUnit.MILLISECONDS.toDays(duration);
if (diffInDays >= 365) {
int year = (int) (diffInDays / 365);
timeDifference = year + mContext.getString(R.string.year_ago);
} else if (diffInDays >= 30) {
int month = (int) (diffInDays / 30);
timeDifference = month + mContext.getString(R.string.month_ago);
}
//if days are not enough to create year then get the days
else if (diffInDays >= 1) {
timeDifference = diffInDays + mContext.getString(R.string.day_ago);
}
//if days value<1 then get the hours
else if (diffInHours >= 1) {
timeDifference = diffInHours + mContext.getString(R.string.hour_ago);
}
//if hours value<1 then get the minutes
else if (diffInMinutes >= 1) {
timeDifference = diffInMinutes + mContext.getString(R.string.min_ago);
}
//if minutes value<1 then get the seconds
else if (diffInSeconds >= 1) {
timeDifference = diffInSeconds + mContext.getString(R.string.sec_ago);
} else if (timeDifference.isEmpty()) {
timeDifference = mContext.getString(R.string.now);
}
return mContext.getString(R.string.added) + " " + timeDifference;
}
其他回答
答案当然是创建一个SimpleDateFormat对象，并使用它来解析“字符串到日期”并将“日期到字符串”格式化。如果您已经尝试了SimpleDateFormat，它没有工作，那么请显示您的代码和您可能收到的任何错误。
附录:String格式中的“mm”与“mm”不一样。用MM表示月，用MM表示分钟。另外，yyyyy和yyyy也不一样。例如,:
import java.text.ParseException;
import java.text.SimpleDateFormat;
import java.util.Date;
public class FormateDate {
public static void main(String[] args) throws ParseException {
String date_s = "2011-01-18 00:00:00.0";
// *** note that it's "yyyy-MM-dd hh:mm:ss" not "yyyy-mm-dd hh:mm:ss"
SimpleDateFormat dt = new SimpleDateFormat("yyyy-MM-dd hh:mm:ss");
Date date = dt.parse(date_s);
// *** same for the format String below
SimpleDateFormat dt1 = new SimpleDateFormat("yyyy-MM-dd");
System.out.println(dt1.format(date));
}
}
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();
}
}
}
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);
SimpleDateFormat dt1 = new SimpleDateFormat("yyyy-mm-dd");
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
