simpledateformat

Convert date string from ISO 8601 format to another

由 送分小仙女□ 提交于 2019-12-25 02:21:17
问题 i have this piece of code , where i am trying to convert a date string from one format to another, and finally i want the date object again. String dateString = "2014-10-04"; SimpleDateFormat oldFormatter = new SimpleDateFormat("yyyy-MM-dd"); Date parsedDate = oldFormatter.parse(dateString); SimpleDateFormat newFormatter = new SimpleDateFormat("dd-MMM-yyyy"); String convertDateStr = newFormatter.format(parsedDate); Date convertedDate = newFormatter.parse(convertDateStr); when i was testing

How to use SimpleDateFormat to show the current Date?

由 做~自己de王妃 提交于 2019-12-25 02:10:01
问题 i'm trying to get the current date and time formatted, this is my code: Date date = new Date(); SimpleDateFormat FORMATTER = new SimpleDateFormat("yyyyMMdd-HHmm"); String formattedDate=""+FORMATTER.format(date); Log.d("XXxxxXxXXXXxxxXXXxXxxxX",formattedDate); Something is going wrong because i'm getting this string: 19700103-0217 That date is incorrect, I should get something like this: 20111118-1217 How to solve this problem? 回答1: Well, I can help you out but I have my own code. Ask me if

SimpleDateFormat ignoring month when parsing

由 自作多情 提交于 2019-12-25 01:49:37
问题 The following code is giving me the parsed date as "Wed Jan 13 00:00:00 EST 2010" instead of "Wed Jun 13 00:00:00 EST 2010". Any ideas much appreciated. SimpleDateFormat sf = new SimpleDateFormat("yyyy-mm-dd'T'HH:mm:ss"); String str = "2010-06-13T00:00:00"; Date date = sf.parse(str); System.out.println(" Date " + date.toString()); 回答1: Try: "yyyy-MM-dd'T'HH:mm:ss" MM means month. mm means minutes. See the documentation for SimpleDateFormat for more details of the supported date and time

Parsing String with spaces to Date, lead to ParseException

由 为君一笑 提交于 2019-12-25 01:18:53
问题 I have date strings in this form Thu Aug 02 00:00:00 GMT+00:00 2012 I have tried to use this method to parse these String in a Date object public Date fromStringToDate(String data) { Date result; SimpleDateFormat sdf = new SimpleDateFormat("yyyy MM dd HH:mm:ss"); try { result = sdf.parse(data); return result; } catch (ParseException e) { e.printStackTrace(); return null; } } But doesn't works and I get this error java.text.ParseException: Unparseable date: "Thu Aug 02 00:00:00 GMT+00:00 2012"

SpringMVC接收String类型时间数值(实体类是Date)转换

由 狂风中的少年 提交于 2019-12-25 00:55:53
SpringMVC接收String类型字符串 bean实体是Date 加入bind @InitBinder public void initBinder(WebDataBinder binder) { SimpleDateFormat dateFormat = new SimpleDateFormat("yyyy-MM-dd"); // SimpleDateFormat dateFormat = new SimpleDateFormat("yyyy-MM-dd'T'HH:mm:ss.SSS Z"); dateFormat.setLenient(false); binder.registerCustomEditor(Date.class, new CustomDateEditor(dateFormat, true)); //true:允许输入空值，false:不能为空值 } 对单独字符串以及实体类bean接收都起效 来源： https://www.cnblogs.com/ukzq/p/12094445.html

SpringFramework的AOP

由 痞子三分冷 提交于 2019-12-24 23:22:27
1、什么是AOP Aspect Oriented Programming， 面向切面编程 2、AOP的作用 方法的增强。 假如有一个代码，里面有10000个方法，现如今需要修改，例：每个方法里面加一个事务。那需要手写10000次吗？肯定不现实。 public class UserServiceImpl implements UserService { @Override public void create ( ) { System . out . println ( "创建一个用户" ) ; } @Override public void delete ( ) { System . out . println ( "删除一个用户" ) ; } } 3、办法 3.1 手动写（非代理） 显然不现实 public class UserServiceImpl2 implements UserService { private static SimpleDateFormat format = new SimpleDateFormat ( "yyyy-MM-dd HH:mm:ss" ) ; @Override public void create ( ) { System . out . println ( "创建一个用户" ) ; System . out . println (

Parsing error for French locale with SimpleDateFormat(string,locale)

由 放肆的年华 提交于 2019-12-24 18:39:27
问题 I have a piece of code like this on my java side: private static DateFormat getHourFormatter(){ //DateFormatSymbols dateFormatSymbols = new DateFormatSymbols(_locale); Locale locale = Locale.FRENCH; //locale : "fr" DateFormat hourFormatter = new SimpleDateFormat( "hh:mm a",locale); //hourFormatter: simpleDateFormat@103068 locale: "fr" hourFormatter.setTimeZone( TimeZone.getTimeZone("GMT") ); return hourFormatter; //hourFormatter: SimpleDateFormat@103068 } protected static boolean

how to get minutes difference between two time in android

由 被刻印的时光 ゝ 提交于 2019-12-24 13:57:46
问题 public int difftime(String string, String string2) { int hours; int min = 0; int days; long difference ; try { SimpleDateFormat simpleDateFormat = new SimpleDateFormat("HH:mm"); Date date1 = simpleDateFormat.parse("08:00 AM"); Date date2 = simpleDateFormat.parse("04:00 PM"); difference = date2.getTime() - date1.getTime(); days = (int) (difference / (1000 * 60 * 60 * 24)); hours = (int) ((difference - (1000 * 60 * 60 * 24 * days)) / (1000 * 60 * 60)); min = (int) (difference - (1000 * 60 * 60

SimpleDateFormat() not giving IllegalArgumentException on passing invalid date format

由 百般思念 提交于 2019-12-24 13:55:03
问题 I passed %%#@-01-01 as date format, but got no exception. Why? SimpleDateFormat df = null; df = new SimpleDateFormat("%%#@-01-01"); 回答1: Quoting the javadoc for SimpleDateFormat: [...] unquoted letters from 'A' to 'Z' and from 'a' to 'z' are interpreted as pattern letters representing the components of a date or time string. Text can be quoted using single quotes (') to avoid interpretation. "''" represents a single quote. All other characters are not interpreted ; they're simply copied into

时间戳比较（在开始时间与结束时间之内）

由 旧城冷巷雨未停 提交于 2019-12-24 11:52:14
时间戳比较（在开始时间与结束时间之内） 实现思路很简单，一般都用它来做登录判断，比如登录的时候检查该用户是否还在有效期范围内，如果不在开始时间与结束时间之内，那么就表明该用户已经过期； 这样的话，首先就可以想起三个参数，第一个需要获取当前系统时间，因为我们登录的时候需要根据当前时间来判断，该用户是否还在有效期之内，然后再用数据库中获取该用户的开始时间与结束时间，这两个字段需要在数据库中存在； 1.首先，我需要需要编写一个uti类，创建一个函数方法： /** * 时间格式化转换 * @param time * @return */ public static long dateToTimestamp(String time) { SimpleDateFormat simpleDateFormat = new SimpleDateFormat("yyyy-MM-dd HH:mm:ss"); try { Date date = simpleDateFormat.parse(time); long ts = date.getTime()/1000; return ts; } catch (ParseException e) { return 0; } } /** * 传入参数返回布尔类型，在此方法进行逻辑判断，如果当前时间（currenttime）在开始时间（startTime）与结束时间
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