Pop*_*orn 2 java datetime parsing date datetime-format
我已经看到了答案,让我在格式化时转换
"14/02/1952 14:52:22"
通过使用DateTimeFormatter
DateTimeFormatter formatter = DateTimeFormat.forPattern("dd/MM/yyyy HH:mm:ss");
DateTime dt = formatter.parseDateTime(string);
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如何将相同的想法应用于此格式的字符串:
"Sunday, September 29, 2013 7:59:58 AM PDT"
我在尝试
DateTimeFormatter formatter = DateTimeFormat.forPattern"EEEE, MMMM dd, yyyy h:mm:ss a zzz";
DateTime dt = formatter.parseDateTime(string);
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但它抱怨这个.
String value = "Sunday, September 29, 2013 7:59:58 AM PDT";
String format = "EEEE, MMMM dd, yyyy h:mm:ss a zzz";
DateTimeFormatter formatter = new DateTimeFormatterBuilder()
.parseCaseInsensitive()
.appendPattern(format)
.toFormatter(Locale.US);
ZonedDateTime zdt = ZonedDateTime.parse(value, formatter);
System.out.println(zdt);
System.out.println(zdt.format(formatter));
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哪个打印
2013-09-29T07:59:58-07:00[America/Los_Angeles]
Sunday, September 29, 2013 7:59:58 AM PDT
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您需要更改预期的格式以满足由此指定的要求 SimpleDateFormat
String value = "Sunday, September 29, 2013 7:59:58 AM PDT";
String format = "EEEE, MMMM dd, yyyy h:mm:ss a zzz";
SimpleDateFormat sdf = new SimpleDateFormat(format);
try {
Date date = sdf.parse(value);
System.out.println(date);
System.out.println(sdf.format(date));
} catch (ParseException ex) {
ex.printStackTrace();
}
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哪个输出......
Mon Sep 30 00:59:58 EST 2013
Sunday, September 29, 2013 7:59:58 AM PDT
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