Let's say I want to represent a person with their name and date of birth. I could write this in Dart:
class Person {
String name;
DateTime dateOfBirth;
}
The problem is that DateTime has a timezone offset so if I want to sort people by age then I need to implement my own Comparator callback function which only looks at the year, month and day properties of dateOfBirth.
An alternative is to add a class invariant to Person to ensure that the dateOfBirth is always in UTC, then my sort comparator becomes much simpler: (p1, p2) => p1.dateOfBirth.compareTo(p2.dateOfBirth)
Both these approaches feel wrong because I don't care about the timezone component - what I really want is a class which just represents (year, month, day) in the Gregorian calendar. I've found this class, but it's not very popular. Is there some standard Dart class that I can use? Or do most Dart programmers just use DateTime to represent such a concept?
Just make sure only year/month/day are included when saving the DoB.
date = getDobFromUser();
dob = DateTime(date.year, date.month, date.day);
If you need more granularity than that, you should include the time they were born and probably store it as milliseconds since the Unix epoch for easy sorting.
Related
I want to get the 4-digit year from today.
I have a variable def todayDate = new Date() in my controller.
I googled to see how to do it. It pointed me to this page.
https://docs.groovy-lang.org/latest/html/groovy-jdk/java/util/Date.html
There is a method called toYear()
Actually, none of the methods in this document works. The Date class in this document is not the same Date class in the controller for sure.
Did Google show me the wrong document? What is the correct way to get the 4-digit year from a Date()?
I was reading and trying to understand all the comments and checking the links. I think I get what is going on here.
When I declare a Date variable in Grails, it is a java.util.Date. If I google how to work with the Date, this gets confusing. The methods that process the Date are not from the java.util.Date class. Majority of the java.util.Date methods are deprecated. The methods you can find from google are actually coming from the org.codehaus.groovy:groovy-dateutil:3.0.9 Groovy date enhancement package.
So, the Date is a java.util.Date but the methods are from the enhancement package.
tl;dr
You said:
I want to get the 4-digit year from today.
String.valueOf( LocalDate.now().getYear() ) // Java syntax.
2022
Avoid legacy classes
The java.util.Date class is terribly flawed. Along with Calendar and the other legacy date-time classes, these were years ago supplanted by the modern java.time classes defined in JSR 310.
Sun, Oracle, and the JCP community all gave up on these legacy classes. I suggest you do the same.
Year
(Forgive the Java syntax as I do not know Groovy.)
If all you want is the year, use Year class.
Year currentYear = Year.now( z ) ;
int y = currentYear.getValue() ;
LocalDate#getYear
For a date-only value, use LocalDate.
Time zone
Determining the current date requires a time zone. For any given moment, the date varies around the globe by time zone.
(Again, Java syntax.)
ZoneId z = ZoneId.of( "America/Edmonton" ) ; // Or, ZoneId.systemDefault() ;
LocalDate today = LocalDate.now( z ) ;
int year = today.getYear() ;
I do not see all these methods in your Groovy API documentation. Perhaps I do not know how that doc works. The classes listed here are built into Java 8 and later.
I want to compare string representations of weeks, e.g. week "01/17" is before "02/17" and after "52/16".
The following code throws an exception, I guess because my string doesn't hint at the exact day of each week. However, I don't care - it could all be Mondays or Thursdays or whatever:
DateTimeFormatter formatter = DateTimeFormatter.ofPattern("ww/YY", Locale.GERMANY);
LocalDate date1 = formatter.parse(str1, LocalDate::from);
Do I need to modify the parser? Or parse to some other format? Unfortunatley there is no object like YearMonth for weeks...
One solution would be to always default to the same day, say the Monday. You could build a custom formatter for that:
DateTimeFormatter fmt = new DateTimeFormatterBuilder()
.appendPattern("ww/YY")
.parseDefaulting(ChronoField.DAY_OF_WEEK, 1)
.toFormatter(Locale.GERMANY);
You can now build LocalDates representing the Monday of the given week:
LocalDate d1 = LocalDate.parse("01/17", fmt);
LocalDate d2 = LocalDate.parse("52/16", fmt);
System.out.println(d1.isAfter(d2));
which prints true because 01/17 is after 52/16.
I wasn't able to find a way for this to work with the DateTimeFormatter class, but I would like to suggest a different approach.
The Threeten Extra library contains a number of classes that were deemed too specific to include in the java.time library. One of them is the YearWeek class you mention.
Your problem can be solved by parsing the week-number and year manually from the input-string and then invoking the YearWeek creator-method like this:
YearWeek yw = YearWeek.of(year, monthOfYear);
tl;dr
YearWeek.parse( "2017-W01" )
ISO 8601
Or parse to some other format?
Yes, use another format.
Use the standard ISO 8601 formats when serializing date-time values to text. The standard includes support for week dates.
For a year-week that would be four year digits, a hyphen, a W, and two digits for the week of the year.
2017-W01
Get clear on your definition of a “week”. The ISO 8601 definition is that:
The Week # 1 contains the first Thursday of the year, and
Runs Monday-Sunday.
So years run either 52 or 53 weeks long. And note that under this definition, the first few days of the year may be in the prior year when week-numbering. Likewise, the last few days of the year may be in the following year when week-numbering.
If you want to indicate a particular day within that week, append a hyphen and a single digit running 1-7 for Monday-Sunday.
Tip: To see ISO 8601 week numbers by default on your computer, you may need to adjust your OS setting. For example, on macOS set System Preferences > Language & Region > Calendar > ISO 8601 to make apps such as Calendar.app to display week numbers with this standard definition.
2017-W01-7
By the way, a couple of similar representations:
An ordinal date meaning the year and the day-of-year-number running from 1-366 is year, a hyphen, and a three-digit number: 2017-123
Month-Day without year is two hyphens, month number, hyphen, and day-of-month number: --01-07
Note that the use of Locale as seen in the Question is irrelevant here with the standard ISO 8601 formats.
YearWeek
Unfortunatley there is no object like YearMonth for weeks...
Ahhh, but there is such a class.
For a class to directly represent the idea of a week-year, see the correct Answer by Henrik. That Answer shows the ThreeTen-Extra library’s class YearWeek.
The YearWeek class can directly parse and generate strings in standard format.
YearWeek yw = YearWeek.parse( "2017-W01" );
You can compare the YearWeek objects with methods: compareTo, equals, isBefore, isAfter.
yw.isBefore( thatYw )
The ThreeTen-Extra project offers other classes such as YearQuarter that you may find useful.
I am using the below query with date filtering, but I am getting wrong result.
SELECT * FROM TRANSACTIONSHISTORY
WHERE DATE > "29-01-2015 12:00:00"
AND DATE < "30-01-2015 00:00:00" AND USERID=abc
I am getting result with date column with value of 29-Jan-2016 records, what am I missing here, can any one help me to get out of this value.
The date format in your SQL will not work because SQLite doesn't have a native datetime type, so it's generally stored either as a string, in YYYY-MM-DD HH:MM:SS.SSS format, or as an numeric value representing the number of seconds since 1970-01-01 00:00:00 UTC. See date and time types on SQLite.org. Note that if you're using the string representation that the sequence is year, month, day (which, when sorting/querying this string field, the this alphanumeric string will sort correctly by year first, then month, and then day, which is critical when doing queries like yours).
If you really stored dates in the database as a string in the DD-MM-YYYY HH:MM:SS format, you should consider changing the format in which you saved the values into one of the approved date formats. It will make the date interactions with the database much, much easier, allowing queries like the one you asked for (though, obviously, with DD-MM-YYYY replaced with YYYY-MM-DD format).
You have cast your string to Date
SELECT * FROM TRANSACTIONSHISTORY WHERE DATE between Datetime('29-01-2015 12:00:00') and Datetime('30-01-2015 00:00:00') AND USERID=abc
The first answer is exactly what you need. What you did in your code would be comparing strings using ASCII values.
I would recommend you to use the linux time stamps like: 1453818208, which is easier to save and compare. In addition, it can always be translated to human-readable dates like: 29-01-2015 12:00:00.
SELECT * FROM TRANSACTIONSHISTORY
WHERE DATE > "29-01-2015 12:00:00"
AND DATE < "30-01-2015 00:00:00" AND USERID=abc
I hope this helps you :)
Try this first try without Time,after that try date and time both , Hope i will work for you
SELECT TRANSACTIONSHISTORY
FROM SHIPMENT
WHERE DATE
BETWEEN '11-15-2010'
AND '30-01-2015'
// you can try this one also
SELECT * FROM TRANSACTIONSHISTORY WHERE DATE BETWEEN "2011-01-11" AND "2011-8-11"
I'm using ActiveRecord's Enum to store the weekdays with the following:
enum weekday: %w(monday tuesday wednesday thursday friday saturday sunday)
When calling the attribute .weekday on the model instance I correctly get the weekday name, e.g. "monday".
How could I get the numeric value (i.e. 0) when I need that instead?
Use my_object[:weekday], or, if you're in the object, just self[:weekday].
UPDATE:
OR (as found by Ms Numbers): .read_attribute_before_type_cast(:weekday)
What you would rather want to do is this:
model_instance.read_attribute(:weekday)
Cleaner, simpler, straight to the point.
I want to store date in my Postgres database.
The only problem is that this date can have optional day or even month.
Example:
User provides time period when he was employed - not necessary full date (day + month + year), but only start year and end year.
However there are users, who worked only from may to october in the same year so month have to be provided too.
How to handle this kind of optional date parts?
Use a proper date type anyway. Do not store text or multiple columns. That would be more expensive and less reliable.
Use the function to_date(), which is fit to deal with your requirements out of the box. For instance, if you call it with a pattern 'YYYYMMDD' and the actual string is missing characters for day, or month and day, it defaults to the first month / day of the year / month:
db=# SELECT to_date('2001', 'YYYYMMDD');
to_date
------------
2001-01-01
db=# SELECT to_date('200103', 'YYYYMMDD');
to_date
------------
2001-03-01
You could store a precision flag indicating year / month / day in addition if you need that.
While the accepted answer is a good one, there is another alternative.
ISO 8601
The ISO 8601 standard defines sensible formats for textual representations of various kinds of date-time values.
A year is represented in the obvious manner, a four-digit number: 2014
A year-month is represented with a required hyphen: 2014-01Note that in other ISO 8601 formats, the hyphen is optional. But not for year month, to avoid ambiguity.
A full date is similar: 2014-08-21 or without optional hyphens: 20140821. I recommend keeping the hyphens.
So you could store the values as text. The length of text would tell you whether it is year-only, year-month, or date.