Parse week strings for comparison using Java 8 - parsing

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.

Related

What Date Class is used in grails?

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.

How do you select all the rows between two values in SQLite?

I'm building an iOS app where I want to retrieve all the values from my database between two dates that the user picks. So for example, I want all the rows from the 1st of March to the 5th of March. Would look something like
SELECT * FROM MAIN WHERE DATE = '01/03/2020' AND ENDS ='05/03/2020'
So from that I would hope to retrieve all data from the 1st,2nd,3rd,4th and 5th of march. Any ideas on how to do this?
Thank you
Try to use comparison operators like:
DATE >= '01/03/2020' AND DATE <= '05/03/2020'
There are two issues:
Date types:
As Datatypes In SQLite Version 3 says:
2.2. Date and Time Datatype
SQLite does not have a storage class set aside for storing dates and/or times. Instead, the built-in Date And Time Functions of SQLite are capable of storing dates and times as TEXT, REAL, or INTEGER values:
TEXT as ISO8601 strings ("YYYY-MM-DD HH:MM:SS.SSS").
REAL as Julian day numbers, the number of days since noon in Greenwich on November 24, 4714 B.C. according to the proleptic Gregorian calendar.
INTEGER as Unix Time, the number of seconds since 1970-01-01 00:00:00 UTC.
Applications can chose to store dates and times in any of these formats and freely convert between formats using the built-in date and time functions.
So storing dates in a dd/MM/yyyy format (using the DateFormatter capitalization convention) is problematic because in the absence of a native date type, it’s going to store them as strings, and therefore all comparisons will be done alphabetically, not chronologically, sorting values like 03/10/2009 (or nonsense strings like 02foobar, for that matter) in between the strings 01/05/2020 and 05/05/2020.
If, however you store them as yyyy-MM-dd, then it just so happens that alphabetical comparisons will yield chronologically correct comparisons, too.
SQL syntax:
Once you have your dates in your database in a format that is comparable, then if you have all of your dates in a single column, you can use the BETWEEN syntax. For example, let’s say you stored all of your dates in yyyy-MM-dd format, then you could do things like:
SELECT * FROM main WHERE date BETWEEN '2020-03-01' AND '2020-03-05';
But needless to say, you can’t use this pattern (or any comparison operators other than equality) as long as your dates are stored in dd/MM/yyyy format.
If you want to show all the data that has values of column "date" between this two dates then:
Select *
from MAIN
where `date` between '01.03.2020' and '05.03.2020';
If you want to show all the data that has values of column "ends" between this two dates then:
Select *
from MAIN
where ends between '01.03.2020' and '05.03.2020';
If you want to show all the data that has values of columns "date" and "ends" between this two dates then:
Select *
from MAIN
where ends between '01.03.2020' and '05.03.2020'
and `date` between '01.03.2020' and '05.03.2020';
Here is a demo

Data retrieving from sqlite DB between two dates - using objective c

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"

Store date with optional month / day

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.

I'm getting "Invalid month in date" trying to run this?

I'm trying to run the following db command against Informix:
delete from table1
where u_id in (select u_id
from table2
where c_id in (select c_id
from ptable
where name = 'Smith'
and dob = '29-08-1946'));
I pass this in as a string to the db.ExecuteNonQuery method in the MS Data Application block and I get the above error?
To get the date format '29-08-1946' to work, you need your DBDATE environment variable set to a value such as "DMY4-" (or "DMY4/"). These are standard variations for the UK (I used them for years; I now use "Y4MD-" exclusively, which matches both ISO 8601:2004 (Date formats) and ISO 9075 (SQL), except when debugging someone else's environment). There are other environment variables that can affect date formatting - quite a lot of them, in fact - but DBDATE takes priority over the others, so it is the big sledgehammer that fixes the problem.
One of the problems is that your notation using a plain string is not portable between US and UK (and ISO) settings of DBDATE. If you have a choice, the neutral constructor for dates is the MDY() function:
WHERE dob = MDY(8,29,1946)
This works regardless of the setting of DBDATE. You can probably use TO_DATE() too:
SELECT TO_DATE('29-08-1946', '%d-%m-%Y') FROM dual;
This generated '1946-08-29 00:00:00.00000' for me - the function generates a DATETIME YEAR TO FRACTION(5) value, but those convert reliably to DATE values in Informix.
You can also use the DATE() function or an explicit cast to DATE (either CAST('29-08-1946' AS DATE) or '29-08-1946'::DATE), but both of those are subject to the whims of the locale of the users.
Your date field is improperly formatted. Since there is no 29th month in the year 1946 that is what is causing the error.
I'd try just swapping the month and day. 08-29-1946.
The way the day and month parts of a date string are read in can depend on your computer's culture settings.
It is always safer to pass date strings to a database in the form 'dd-MMM-yyyy' (i.e. '29-aug-1946')
It's even safer to pass them as YYYY-MM-DD, the dd-MMM-yyyy in that example will fail on a server with a (for example) French locale.

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