Monday, July 25, 2011

[java] Java Simple Date Format 速查表(for SimpleDateFormat)

假設我要把今天的日期轉換成以下其中一個格式該如何做呢?

首先在程式碼裡面先用Calendar取得現在的時間與日期:
Calendar today = Calendar.getInstance();

但是取得的today格式很難閱讀,所以再轉換成date格式:
Date date = today.getTime();

接著來做時間格式的轉換:
DateFormat df = new SimpleDateFormat("yyyy-MM-dd");//可將括號內的字串換成以下格式
String dateFormat = df.format(date);
最後即可取得需要的時間格式囉!

LetterDate or Time Component Presentation Examples
G Era designator Text AD
yyyy,yy *Year Year 1996; 96
MMMM;MMM;MM;M *Month in year Month July; Jul; 07;7
w Week in year Number 27
W Week in month Number 2
D Day in year Number 189
dd;d *Day in month Number 05;5
F Day of week in month Number 2
EEEE;EEE;E Day in week Text Tuesday; Tue; 2
a Am/pm marker Text PM
HH;H *Hour in day (0-23) Number 02;2
kk;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
mm;m *Minute in hour Number 08;8
ss;s *Second in minute Number 05;5
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

Ex : 2001-07-04 12:08:56
Date and Time Pattern Result
"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

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