I have the following String representation of 24hr like 1423 and 1956
I would like to check if the current time (with time zone) is between this pair
my current solution gets the hour and minute of the day using
int h = Instant.now().atZone(ZoneId.systemDefault()).getHour()
int t = Instant.now().atZone(ZoneId.systemDefault()).getMinute()
creating a 4 digit strings from the numbers, and do a string comparison
I don't like my solution at all, What's the elegant way to do it?
You can convert String representation of time to LocalTime as follows:
var formatter = DateTimeFormatter.ofPattern("HHmm");
var lowerTime = LocalTime.parse("1423", formatter);
var upperTime = LocalTime.parse("1956", formatter);
Then you can use isAfter()/isBefore methods to check if it is in between:
var nowTime = LocalTime.now(ZoneId.systemDefault());
boolean inBetween = nowTime.isAfter(lowerTime) && nowTime.isBefore(upperTime);
Create a LocalTime object using the ZonedDateTime.now() reference and subsequent call to ZonedDateTime#toLocalTime. Then create a DateTimeFormatter object using the Pattern HHmm which stands for hours (HH) and minutes (mm). Then you can create a LocalTime from the start time 1423 using the pattern, and an end time of 1956 from the pattern. You can then create a boolean representing whether or not your time is between by determining if the current time is after the start time and before the end time.
LocalTime time = ZonedDateTime.now().toLocalTime();
DateTimeFormatter pattern = DateTimeFormatter.ofPattern("HHmm");
LocalTime startTime = LocalTime.parse("1423", pattern);
LocalTime endTime = LocalTime.parse("1956", pattern);
boolean between = time.isAfter(startTime) && time.isBefore(endTime);
As you note in the question, your strings are in 24-hour time with leading zeros, so you can compare them as strings. Here is a more convenient way to convert the current time to a string in the required format:
String now = DateTimeFormatter.ofPattern("HHmm")
.withZone(ZoneId.systemDefault())
.format(Instant.now());
Related
I have:
a joda LocalDate, so it has no time information and just date
a string which contains time and zone information, like "14:20 CEST"
Either of them can be absent (Scala's Option).
How can I combine these two to get joda LocalDateTime, i.e. entity representing only date and time with no timezone?
To combine these 2 options, the natural way is to use flatMap method like this:
val onlyDateOption: Option[LocalDate] = ???
val timeAndZoneOption: Option[String] = ???
val result: Option[LocalDateTime] = onlyDateOption.flatMap { onlyDate =>
timeAndZoneOption.map { timeAndZone =>
// Some logic here to build the LocalDateTime from onlyDate and timeAndZone
}
}
Which can also be written with for-comprehension in more readible way:
val result: Option[LocalDateTime] = for {
onlyDate <- onlyDateOption
timeAndZone <- timeAndZoneOption
} yield {
// Some logic here to build the LocalDateTime from onlyDate and timeAndZone
}
Now, how to use Joda to build what you're expecting can probably be done in various different ways, one could be:
onlyDate
.toLocalDateTime(LocalTime.MIDNIGHT)
.withHourOfDay(...) // hour extracted from the string somehow
.withMinuteOfHour(...) // minute extracted from the string somehow
I'm not familiar with Joda API, there is probably another easier way
How to combine LocalDate and String using Joda-Time
You have already got an answer treating the use of Option in detail. Here I want to go into more detail with the combination of your LocalDate and your String into a LocalDateTime using Joda-Time. I understand that you are getting a Joda-Time LocalDate from legacy code and need to return a Joda-Time LocalDateTime to legacy code. I am assuming that you know the time zone the abbreviation of which is in the string. I think that you should validate that abbreviation since Central European Time uses the abbreviation CET during the standard time part of the year and CEST during summer time (DST). Excuse my Java code.
DateTimeUtils.setDefaultTimeZoneNames(createTimeZoneNamesMap());
DateTimeFormatter timeFormatter = DateTimeFormat.forPattern("H:mm z");
LocalDate date = new LocalDate(2021, 5, 22);
String timeAndZoneString = "14:20 CEST";
LocalTime time = LocalTime.parse(timeAndZoneString, timeFormatter);
DateTime dateTime = date.toDateTime(time, ZONE);
// Validate time zone abbreviation; take overlap at fall-back into account
String earlierCorrectTimeString = dateTime.withEarlierOffsetAtOverlap()
.toString(timeFormatter);
if (! timeAndZoneString.equals(earlierCorrectTimeString)) {
String laterCorrectTimeString = dateTime.withLaterOffsetAtOverlap()
.toString(timeFormatter);
if (! timeAndZoneString.equals(laterCorrectTimeString)) {
throw new IllegalStateException("Incorrect time zone abbreviation for date");
}
}
LocalDateTime ldt = dateTime.toLocalDateTime();
System.out.println(ldt);
Output:
2021-05-22T14:20:00.000
I have used these two auxiliary declaration:
private static final DateTimeZone ZONE = DateTimeZone.forID("Europe/Paris");
private static Map<String, DateTimeZone> createTimeZoneNamesMap() {
Map<String, DateTimeZone> names = new HashMap<>(4);
names.put("CET", ZONE);
names.put("CEST", ZONE);
return names;
}
The validity of the time on the date is also validated: date.toDateTime() validates that the resulting DateTime would not fall in the gap at the spring-forward and throws an IllegalInstantException: if it would.
If the hours you receive in the string are always two digits, the format pattern string needs to specify this, so HH:mm z.
Please be aware that you are losing information in a corner case: If the time falls in the overlap at fall-back, the time zone abbreviation disambiguates, but the LocalDateTime that you produce is ambiguous. For example the date is 2021-10-31 and the time string is 2:20 CEST. Then we know that the time is in the summer time part of the year, that is, before the clocks are turned back. You return 2021-10-31T02:20:00.000, and the receiver won’t be able to tell whether to understand it as 2021-10-31T02:20:00.000+02:00 (summer time) or 2021-10-31T02:20:00.000+01:00 (standard time).
my method accepts - hours, minutes, seconds and milliseconds separated by sign / as a string parameter
how can I add to the current date the parameters that come to the method.
Example 1: today, 02/10/2021, the method receives metnode data (10/10/10/10) - output - 02/10/2021 10:10:10
Example 2: today, 02/10/2021, the method receives metnode data (55/10/10/10) - output - 02/12/2021 07:10:10
That is, you need to add 55 hours 10 seconds 10 seconds and 10 milliseconds to the current date.
you cannot use the Calendar and StringTokenizer classes.
public void method(String s) {
s = s.replaceAll("/", "-");
DateTimeFormatter formatter = DateTimeFormatter.ofPattern("dd.MM.yyyy HH:mm:ss");
final LocalDateTime now = LocalDateTime.parse(s, formatter.withResolverStyle(ResolverStyle.LENIENT));
System.out.println(now);
}
i found the withResolverStyle (ResolverStyle.LENIENT) method
but did not understand how to use it.
A lenient DateTimeFormatter is enough
I don’t know if it’s the best solution. That probably depends on taste. It does use the ResolverStyle.LENIENT that you mentioned and generally works along the lines of the code in your question, only fixed and slightly simplified.
My formatter includes both date and time. This is necessary for surplus hours to be converted to days.
private static final DateTimeFormatter formatter = new DateTimeFormatterBuilder()
.appendPattern("uuuu-MM-dd H/m/s/")
.appendValue(ChronoField.MILLI_OF_SECOND)
.toFormatter()
.withResolverStyle(ResolverStyle.LENIENT);
Next thing we need a string that matches the formatter. So let’s prepend the date to the time string that we already have got:
String timeString = "55/10/10/10";
LocalDate today = LocalDate.now(ZoneId.of("America/Regina"));
String dateTimeString = "" + today + ' ' + timeString;
LocalDateTime dateTime = LocalDateTime.parse(dateTimeString, formatter);
System.out.println(dateTime);
The output from my code when I ran it today (February 10) was:
2021-02-12T07:10:10.010
A different idea: use Duration
Edit: An alternative is to use the Duration class. A reason for doing that would be that it really appears that you are adding a duration rather than setting the time of day. A liability is that parsing your string into a Duration is a bit tricky. The Duration.parse method that we want to use only accepts ISO 8601 format. It goes like PT55H10M10.010S for a period of time of 55 hours 10 minutes 10.010 seconds. And yes, milliseconds need to be given as a fraction of the seconds.
String isoTimeString = timeString.replaceFirst("(/)(\\d+)$", "$100$2")
.replaceFirst("(\\d+)/(\\d+)/(\\d+)/0*(\\d{3})", "PT$1H$2M$3.$4S");
Duration dur = Duration.parse(isoTimeString);
LocalDateTime dateTime = LocalDate.now(ZoneId.of("Asia/Kathmandu"))
.atStartOfDay()
.plus(dur);
When I ran it just now — already February 11 in Kathmandu, Nepal — the output was:
2021-02-13T07:10:10.010
I am using two calls to replaceFirst(), each time using a regular expression. The first call simply adds some leading zeroes to the milliseconds. $1 and $2 in the replacement string give us what was matched by the first and the second group denoted with round brackets in the regular expression.
The second replaceFirst() call established the ISO 8601 format, which includes making sure that the milliseconds are exactly three digits so they work as a decimal fraction of the seconds.
Link: ISO 8601
Try this:
public void method(String s) {
String[] arr = s.split("/");
LocalDateTime now = LocalDateTime.of(
LocalDate.now(), LocalTime.of(0, 0))
.plusHours(Integer.parseInt(arr[0]))
.plusMinutes(Integer.parseInt(arr[1]))
.plusSeconds(Integer.parseInt(arr[2]))
.plusNanos(Integer.parseInt(arr[3]) * 1_000_000L);
DateTimeFormatter formatter = DateTimeFormatter.ofPattern("MM/dd/yyyy HH:mm:ss");
System.out.println(now.format(formatter));
}
Look into the LocalDateTime documentation. It offers various means for combining dates. Such as:
plus(amount, unit)
plusDays(days)
plusHours(hours)
plusMinutes(minutes)
just for simplicity , you can your LocalDateTime class. it is easy to understand. please refer to below code is used to add the hours, minuts, second and nanos to current Date Time.
this Date Time then can easy formatted by any format pattern as required.
public void addDateTime(int hours, int minuts, int seconds, int nanos) {
LocalDateTime adt = LocalDateTime.now();
System.out.println(adt);
adt = adt.plusHours(hours);
adt = adt.plusMinutes(minuts);
adt = adt.plusSeconds(seconds);
adt = adt.plusNanos(nanos);
System.out.println(adt);
}
I have a long value that represents a datetime like this 20200319234500 (translates into March 19th, 2020 11:45:00PM)
I want this long value (20200319234500) converted into another timezone again in long format so I can do a greater than and less than comparison with the current date time in local timezone.
I want to do it efficiently so I dont have to create any objects during run time or do string creations after the start up.
but it looks like I must first convert long time to a string and then call ZonedDateTime.parse() function to get a datetime and then do a comparison. Is there another way of doing this?
//This is known at compile time
ZoneId baseZone = ZoneId.of("Europe/Paris");
//This is known at compile time
ZoneId localZone = ZoneId.of("America/New_York");
//This is known at compile time
DateTimeFormatter formatter = DateTimeFormatter.ofPattern("uuuuMMddHHmmss").withZone(baseZone);
long baseDateTime = 20210321234500L;
//Dont want o be doing string operations. Is there a way to keep it in long and get another long in a different zone?
ZonedDateTime convertedBaseDateTime = ZonedDateTime.parse(Long.toString(baseDateTime), formatter);
//Can I just get a long that represents current time in local zone? local zone is not JVM zone
ZonedDateTime localDateTime = ZonedDateTime.now(localZone);
//Thats the only operation I want to perform
boolean result = convertedBaseDateTime.isBefore( localDateTime);
You can do some maths to get the year, month, day, hour. minute, second from the long, and then you can pass it to ZonedDateTime.of
long baseDateTime = 20210321234500L;
int year = (int)(baseDateTime / 10000000000L);
int month = (int)(baseDateTime / 100000000L % 100);
int day = (int)(baseDateTime / 1000000L % 100);
int hour = (int)(baseDateTime / 10000L % 100);
int minute = (int)(baseDateTime / 100L % 100);
int second = (int)(baseDateTime % 100);
ZonedDateTime convertedBaseDateTime = ZonedDateTime.of(year, month, day, hour, minute, second, 0, baseZone);
This won't create new strings.
After that, notice that if you just want to check if a date time is before "now", you don't need a zone for "now". You just need to compare the numbers of (milli/nano)seconds since the epoch of then, and now.
// don't need these
// ZoneId localZone = ZoneId.of("America/New_York");
// ZonedDateTime localDateTime = ZonedDateTime.now(localZone);
// you just need to compare
convertedBaseDateTime.toEpochSecond() * 1000 < System.currentTimeMillis()
That said, if performance is so important for you, maybe you shouldn't use Java, and should instead use a more low-level language.
long baseDateTime = 20210321234500L;
LocalDateTime time=LocalDateTime.ofEpochSecond(baseDateTime /1000,0,ZoneOffset.ofHours(8));
//then you can use time2.isAfter() or some other methond to comparable that is link to jdk 8 API .
I have a map of string values which represent down times for different components.
dependencyMap.put ("sut", "14:26:12,14:27:19,00:01:07;15:01:54,15:02:54,00:01:00;15:44:30,15:46:30,00:02:00;16:10:30,16:11:30,00:01:00");
dependencyMap.put ("jms", "14:26:12,14:28:12,00:02:00;15:10:50,15:12:55,00:02:05;15:42:30,15:43:30,00:01:00;16:25:30,16:27:30,00:02:00");
The strings represent the start, end and duration of down times.
(start)14:26:12,(end)14:27:19,(duration)00:01:07
I read the values in, then add them to a list of DependencyDownTime objects which hold the Long values startTime, endTime and duration.
jArray.forEach (dependency ->{
String downTimeValues = knownDowntimesMap.get(dependency);
final String[] downtime = downTimeValues.split (";");
for (final String str : downtime) {
final DependencyDownTime depDownTime = new DependencyDownTime ();
final String[] strings = str.split (",");
if (strings.length == 3) {
final DateFormat dateFormat = new SimpleDateFormat ("HH:mm:ss");
try {
depDownTime.setStartTime(dateFormat.parse (strings[0]).getTime ());
depDownTime.setEndTime (dateFormat.parse (strings[1]).getTime ());
depDownTime.setDuration (dateFormat.parse (strings[2]).getTime ());
downTimes.add (depDownTime);
} catch (final ParseException e) {
//logger.warn (e.getMessage (), e);
}
} else {
//logger.warn ("");
}
}
I then perform simple arithmetic on the values, which calculates the total down time for each component.
// sort the list by start time
Collections.sort(downTimes, Comparator.comparing (DependencyDownTime::getStartTime));
int i = 1;
Long duration = 0L;
for(DependencyDownTime dts: downTimes){
Long curStart = dts.getStartTime ();
Long curEnd = dts.getEndTime();
Long nextStart = downTimes.get(i).getStartTime ();
Long nextEnd = downTimes.get(i).getEndTime ();
if(duration == 0){
duration = dts.getDuration();
}
if(curStart.equals(nextStart) && curEnd < nextEnd){
duration += (nextEnd - curEnd);
}
else if(nextStart > curEnd){
duration += downTimes.get(i).getDuration();
}
else if( curStart < nextStart && curEnd > nextStart){
duration += (nextEnd - curEnd);
}
else if(curEnd == nextStart){
duration += downTimes.get(i).getDuration();
}
i++;
if(i == downTimes.size ()){
componentDTimeMap.put (application, duration);
return;
}
The expected values should be something like 1970-01-01T 00:14:35 .000+0100, a matter of minutes. The actual result is usually extremely high off by a matter of hours in the difference 1969-12-31T 15:13:35 .000+0100
I have 2 questions.
Am I parsing the values correctly?
If my calculations are a little off when adding and subtracting the long values. When I convert the values back to Date format will there be a drastic difference in the expected value?
As explained in your other question, don't mistake those 2 different concepts:
a time of the day: it represents a specific point of a day, such as 10 AM or 14:45:50
a duration: it represents an amount of time, such as "1 hour and 10 minutes" or "2 years, 3 months and 4 days". The duration doesn't tell you when it starts or ends ("1 hour and 10 minutes" relative to what?), it's not attached to a chronology, it doesn't correspond to a specific point in the timeline. It's just the amount of time, by itself.
In your input, you have:
(start)14:26:12,(end)14:27:19,(duration)00:01:07
The start and end represents times of the day, and the duration represents the amount of time. SimpleDateFormat is designed to work with dates and times of the day, but not with durations. Treating the duration as a time of the day might work, but it's a hack as explained in this answer.
Another problem is that when SimpleDateFormat parses only a time, it defaults the day to January 1st 1970 at the JVM default timezone, leading to all the strange results you see. Unfortunately there's no way to avoid that, as java.util.Date works with full timestamps. A better alternative is to use the new date/time API.
As in your other question you're using Java 8, I'm assuming you can also use it here (but if you're using Java <= 7, you can use the ThreeTen Backport, a great backport for Java 8's new date/time classes. The only difference is the package names (in Java 8 is java.time and in ThreeTen Backport (or Android's ThreeTenABP) is org.threeten.bp), but the classes and methods names are the same).
As you're working only with times, there's no need to consider date fields (day/month/year), we can use a LocalTime instead. You can parse the strings directly, because they are in ISO861 compliant format:
LocalTime start = LocalTime.parse("14:26:12");
LocalTime end = LocalTime.parse("14:27:19");
Unfortunately there are no built-in parsers for a duration, so you'll have to parse it manually:
// parse the duration manually
String[] parts = "00:01:07".split(":");
Duration d = Duration
// get hours
.ofHours(Long.parseLong(parts[0]))
// plus minutes
.plusMinutes(Long.parseLong(parts[1]))
// plus seconds
.plusSeconds(Long.parseLong(parts[2]));
Another alternative is to remove the durations from your input (or ignore them) and calculate it using the start and end:
Duration d = Duration.between(start, end);
Both will give you a duration of 1 minute and 7 seconds.
My suggestion is to change the DependencyDownTime to store start and end as LocalTime objects, and the duration as a Duration object. With this, your algorithm would be like this:
Duration total = Duration.ZERO;
for (...) {
LocalTime curStart = ...
LocalTime curEnd = ...
LocalTime nextStart = ...
LocalTime nextEnd = ...
if (total.toMillis() == 0) {
duration = dts.getDuration();
}
if (curStart.equals(nextStart) && curEnd.isBefore(nextEnd)) {
total = total.plus(Duration.between(curEnd, nextEnd));
} else if (nextStart.isAfter(curEnd)) {
total = total.plus(downTimes.get(i).getDuration());
} else if (curStart.isBefore(nextStart) && curEnd.isAfter(nextStart)) {
total = total.plus(Duration.between(curEnd, nextEnd));
} else if (curEnd.equals(nextStart)) {
total = total.plus(downTimes.get(i).getDuration());
}
i++;
if (i == downTimes.size()) {
// assuming you want the duration as a total of milliseconds
componentDTimeMap.put(application, total.toMillis());
return;
}
}
You can either store the Duration object, or the respective value of milliseconds. Don't try to transform it to a Date, because a date is not designed nor supposed to work with durations. You can adapt this code to format a duration if you want (unfortunately there are no native formatters for durations).
Limitations
The code above assumes that all start and end times are in the same day. But if you have start at 23:50 and end at 00:10, should the duration be 20 minutes?
If that's the case, it's a little bit trickier, because LocalTime is not aware of the date (so it considers 23:50 > 00:10 and the duration between them is "minus 23 hours and 40 minutes").
In this case, you could do a trick and assume the dates are all at the current date, but when start is greater than end, it means that end time is in the next day:
LocalTime start = LocalTime.parse("23:50");
LocalTime end = LocalTime.parse("00:10");
// calculate duration
Duration d;
if (start.isAfter(end)) {
// start is after end, it means end is in the next day
// current date
LocalDate now = LocalDate.now();
// start is at the current day
LocalDateTime startDt = now.atTime(start);
// end is at the next day
LocalDateTime endDt = now.plusDays(1).atTime(end);
d = Duration.between(startDt, endDt);
} else {
// both start and end are in the same day
// just calculate the duration in the usual way
d = Duration.between(start, end);
}
In the code above, the result will be a Duration of 20 minutes.
Don't format dates as durations
Here are some examples of why SimpleDateFormat and Date aren't good to handle durations of time.
Suppose I have a duration of 10 seconds. If I try to transform it to a java.util.Date using the value 10 to a date (AKA treating a duration as a date):
// a 10 second duration (10000 milliseconds), treated as a date
Date date = new Date(10 * 1000);
System.out.println(date);
This will get a date that corresponds to "10000 milliseconds after unix epoch (1970-01-01T00:00Z)", which is 1970-01-01T00:00:10Z. But when I print the date object, the toString() method is implicity called (as explained here). And this method converts this millis value to the JVM default timezone.
In the JVM I'm using, the default timezone is America/Sao_Paulo, so the code above outputs:
Wed Dec 31 21:00:10 BRT 1969
Which is not what is expected: the UTC instant 1970-01-01T00:00:10Z corresponds to December 31st 1969 at 9 PM in São Paulo timezone.
This happens because I'm erroneously treating the duration as a date (and the output will be different, depending on the default timezone configured in the JVM).
A java.util.Date can't (must not) be used to work with durations. Actually, now that we have better API's, it should be avoided whenever possible. There are too many problems and design issues with this, just don't use it if you can.
SimpleDateFormat also won't work properly if you handle the durations as dates. In this code:
SimpleDateFormat dateFormat = new SimpleDateFormat("HH:mm:ss");
Date d = dateFormat.parse("10:00:00");
The input has only time fields (hour, minute and second), so SimpleDateFormat sets the date to January 1st 1970 at the JVM default timezone. If I System.out.println this date, the result will be:
Thu Jan 01 10:00:00 BRT 1970
That's January 1st 1970 at 10 AM in São Paulo timezone, which in UTC is equivalent to 1970-01-01T13:00:00Z - so d.getTime() returns 46800000.
If I change the JVM default timezone to Europe/London, it will create a date that corresponds to January 1st 1970 at 10 AM in London (or UTC 1970-01-01T09:00:00Z) - and d.getTime() now returns 32400000 (because 10 AM in London and 10 AM in São Paulo happened at different instants).
SimpleDateFormat isn't the right tool to work with durations - it isn't even the best tool to work with dates, actually.
I am trying to calculate the difference between two times, which are represented as longs in the Format HHmm 24 hour time. E.g 4:30pm is represented by the long 0430.
I am happy for the difference to be in minutes.
Is there a simple calculation that can be done to achieve this? I am aware of Java's Date class, however I want to avoid having to store dummy date information just for a calculation on time.
Thanks!
Putting aside the fact that this is a really, really bad way to store times, the easiest way to do this is to convert the HHMM time to minutes since the start of the day:
long strangeTimeFormatToMinutes(long time) {
long minutes = time % 100;
long hours = time / 100;
return minutes + 60 * hours;
}
Then just use plain old subtraction to get the difference.
You may also want to add validation that minutes and hours are in the ranges you expect, i.e. 0-59 and 0-23.
You mentioned that you didn't want to use the Date class because it required you to use a dummy date. The LocalTime class does not require that.
LocalTime start = LocalTime.of(6,15,30,200); // h, m, s, nanosecs
LocalTime end = LocalTime.of(6,30,30,320);
Duration d = Duration.between(start, end);
System.out.println(d.getSeconds()/60);
Pad zeros
First convert your integer to a 4-character string, padding with leading zeros.
For example, 430 becomes 0430 and parsed as 04:30. Or, 15 becomes 0015 and parsed as quarter past midnight, 00:15.
String input = String.format( "%04d", yourTimeAsInteger );
LocalDate
The LocalTime class represents a time-of-day value with no date and no time zone.
DateTimeFormatter f = DateTimeFormatter.ofPattern( "HHmm" );
LocalTime ld = LocalTime.parse( input , f ) ;