You will always have problems if you try to roll your own date/time arithmetic. Use the functions provided by the JDK, java.util.Calendar and the new java.time package, or a library like Joda Time. Try: Period.between(date-of-birth , today’s date). 1 solved Measure difference between dates [closed]
This should work as you specified in your question. DateFormat dateFormat = new SimpleDateFormat(“yy-MMM-dd HH:mm:ss a”); System.out.println(dateFormat.format(new Date())); The date format you have in your question is something like: DateFormat dateFormat = new SimpleDateFormat(“yyyyMMddHmmss”); 2 solved Convert String to Date object [duplicate]
You should be using SimpleDateFormat(“dd-MM-yyyy”). DD is for day in year, like there are 365 days in a year. 3 solved SimpleDateFormat returning incorrect day on given date [closed]
You’re comparing strings. That compares their characters, one by one from left-to-right, until it finds a difference, and then uses that difference as the result. Since “2” is > “0”, that string is greater than the other. You need to parse the dates and compare the result. Do not just use new Date(dateFormat) or similar, … Read more
Check this out: try { SimpleDateFormat format1 = new SimpleDateFormat(“MMM dd, yyyy”); SimpleDateFormat format2 = new SimpleDateFormat(“yyyy-MM-dd”); String stringDate = “Dec 13, 2013”; Date date = format1.parse(stringDate); System.out.println(format2.format(date)); } catch (ParseException exp) { // Take appropriate action } Output: 2013-12-13 3 solved How to change the format of date in java [duplicate]
In the spirit of teaching, see comments: var stopDate = new Date(“2014-09-04”); // One day AFTER the last date we care // about var results = []; // An array for our results, initially // empty var date = new Date(“2014-08-28”); // Our starting date var entry; // The “current” month’s entry in the loop … Read more
This is logic that you would normally put inside your PHP script, not smarty. Try something like this: empty($_GET[‘day’]) ? $day = date(‘l’) : $day = $_GET[‘day’]; $date = strtotime(“last {$day}”); echo “News from “.date(“l, F j”, $date); This script assumes that you pass the day as a GET parameter, for example index.php?day=monday. You can … Read more
Using SimpleDateFormatter you can convert from date string to date and date to date String public final class DateUtil { private static final String DEFAULT_DATE_FORMAT = “dd-MMM-yyyy”; private DateUtil() { } public static final String formatDate(final Date date, final String dateFormat) { DateFormat format = new SimpleDateFormat(dateFormat, Locale.ENGLISH); return format.format(date); } public static final Date … Read more
$date = “2017-12-31”; echo date(“Y-m-d”, strtotime($date . ” +5 days”)); Strtotime() can translate some sentences to actions. See example 1 solved Add +1 or more day to DATETIME in php eaven if that day is in a different month [closed]
In R, using dplyr: data %>% group_by(Name, CoordinateX, CoordinateY) %>% arrange(desc(Date)) %>% distinct() %>% ungroup() Give the output: Name Date CoordinateX CoordinateY Aaa 2018-08-29 650000 134999 Bbb 2010-08-29 650000 134999 Bbb 2010-08-29 655600 134999 Ccc 2010-08-29 655600 134999 solved Remove duplicate on multiple columns keep newest [closed]
I got this like following way $mydate=getdate(date(“U”)); echo “$mydate[weekday], $mydate[month] $mydate[mday], $mydate[year]”; OR $today = date(DATE_RFC822); echo date(‘l,F d,Y’,strtotime($today)); solved How to get date in Thursday, September 26, 2013 format? [closed]
You are experiencing SimpleDateFormat behaving as designed. It’s a behaviour that comes as a negative surprise to most. There are two solutions: the recommended one and the discouraged one. Recommended solution: LocalTime DateTimeFormatter timeFormat = DateTimeFormatter.ofPattern(“hh:mm a”, Locale.ROOT); try { LocalTime lt = LocalTime.parse(startTime, timeFormat); startTime = lt.format(timeFormat); System.out.println(startTime); } catch (DateTimeParseException e) { System.out.println(“Not … Read more
Assuming that seconds is a unix epoch (UTC), you should just use function date_str(seconds) { var dt = new Date(seconds*1000); console.log(dt); … instead. The get…() methods will respect the local timezone. If you don’t want that, you should use the getUTC…() equivalents. 0 solved How to make JS date respect local timezone?
The Date Javadoc notes A year y is represented by the integer y – 1900. A month is represented by an integer from 0 to 11; 0 is January, 1 is February, and so forth; thus 11 is December. To get the correct Date with the deprecated Date constructor, it would need to be something … Read more
declare @timediff table(id int identity(1,1),Name varchar(50),logtime datetime) insert into @timediff Select Name,logtime from timediff declare @datetimeDiff int declare @i int=1 while(@i<=(Select count(*) from @timediff)) Begin Select @datetimeDiff=datediff(hour,logtime, (Select logtime from @timediff where id=@i+1)) from @timediff where id=@i if(@datetimeDiff>=1) BEGIN Select @datetimeDiff –You can write your update code here END Set @i=@i+2 END 0 solved Need … Read more