I am having struggle to convert a number format to a date/datetime format in PowerBI ( in the query editor).
The rowformat in csv looks like this (german timezone):
202007201155180
201912012200001
which would lead to:
20.07.2020 11:55:18
01.12.2019 22:20:00
You can use division (Number.IntegerDivide) and modulo (Number.Mod) to calculate the date parts (year, month, day, hour, etc.) and construct a datetime value (#datetime) in a custom column, e.g. like this:
#datetime(
Number.IntegerDivide([num], 100000000000), // Year
Number.IntegerDivide(Number.Mod([num], 100000000000), 1000000000), // Month
Number.IntegerDivide(Number.Mod([num], 1000000000), 10000000), // Day
Number.IntegerDivide(Number.Mod([num], 10000000), 100000), // Hour
Number.IntegerDivide(Number.Mod([num], 100000), 1000), // Minute
Number.IntegerDivide(Number.Mod([num], 1000), 10) // Second
)
This will give you a new datetime column:
But note, that 201912012200001 is not 01.12.2019 12:20:00 as you said, but 01.12.2019 22:00:00, and the last digit is not converted, so the fraction of the second (tenth of the second) is lost.
Related
I have a column called PairDt, a string that contains a date value in the last 5 characters. I want to compare that date value with the date value in the Day column, which contains dates in the YYYY-MM-DD format.
PairDt Day
----------------------------------
DCS-CNY-Yunbi-42606 2016-08-24
DCS-CNY-Yunbi-42607 2016-08-25
DCS-CNY-Yunbi-42608 2016-08-26
DCS-CNY-Yunbi-42609 2016-08-27
DCS-CNY-Yunbi-42610 2016-08-28
How do I convert Day to a value?
I'm trying to isolate Date values in PairDt that does not match the date value in Days
This 5 digit number at the end of PairDt looks like number of days since December 30th 1899. To convert this number to date use DATEADD to add as many days. To convert a date to number, use DATEDIFF to calculate the number of days. Something like this code:
declare #PairDt varchar(50) = 'DCS-CNY-Yunbi-42606', #Day date = '2016-08-24'
select DATEADD(d, cast(right(#PairDt, 5) as int), '1899-12-30'), DATEDIFF(day, '1899-12-30', #Day)
I'm trying to manipulate a date value to go back in time exactly 1 ISO-8601 year.
The following does not work, but best describes what I want to accomplish:
date_add(date '2018-01-03', interval -1 isoyear)
I tried string conversion as an intermediate step, but that doesn't work either:
select parse_date('%G%V%u',safe_cast(safe_cast(format_date('%G%V%u',date '2018-01-03') as int64)-1000 as string))
The error provided for the last one is "Failed to parse input string "2017013"". I don't understand why, this should always resolve to a unique date value.
Is there another way in which I can subtract an ISO year from a date?
This gives the corresponding day of the previous ISO year by subtracting the appropriate number of weeks from the date. I based the calculation on the description of weeks per year from the Wikipedia page:
CREATE TEMP FUNCTION IsLongYear(d DATE) AS (
-- Year starting on Thursday
EXTRACT(DAYOFWEEK FROM DATE_TRUNC(d, YEAR)) = 5 OR
-- Leap year starting on Wednesday
(EXTRACT(DAY FROM DATE_ADD(DATE(EXTRACT(YEAR FROM d), 2, 28), INTERVAL 1 DAY)) = 29
AND EXTRACT(DAYOFWEEK FROM DATE_TRUNC(d, YEAR)) = 4)
);
CREATE TEMP FUNCTION PreviousIsoYear(d DATE) AS (
DATE_SUB(d, INTERVAL IF(IsLongYear(d), 53, 52) WEEK)
);
SELECT PreviousIsoYear('2018-01-03');
This returns 2017-01-04, which is the third day of the 2017 ISO year. 2018-01-03 is the third day of the 2018 ISO year.
I creating an application, for that I need to find data by month using JPA and java.time.LocalDate. So, is it possible to retrieve data by month from mysql?
Thanks in advance for help.
First find start and end date of month and use between method of JPA to find data of current month.
LocalDate start = LocalDate.ofEpochDay(System.currentTimeMillis() / (24 * 60 * 60 * 1000) ).withDayOfMonth(1);
LocalDate end = LocalDate.ofEpochDay(System.currentTimeMillis() / (24 * 60 * 60 * 1000) ).plusMonths(1).withDayOfMonth(1).minusDays(1);
In Repository
List<Object> findByCreatedateGreaterThanAndCreatedateLessThan(LocalDate start,LocalDate end);
Its better to use the between keyword, it makes things allot shorter.
List<Object> findByCreatedateBetween(LocalDate start,LocalDate end);
Also if you want to use the LocalDate or LocalDateTime objects with Spring Data you should use the converter class Jsr310JpaConverters or else the documents will be stored as Blobs instead of Dates (which is bad for portability of the database). Please see this tutorial on how to implement the Converter.
https://www.mkyong.com/spring-boot/spring-boot-spring-data-jpa-java-8-date-and-time-jsr310/
tl;dr
YearMonth.now( ZoneId.of( "Pacific/Auckland" ) ) // Get current month for particular time zone.
.atDayOfMonth( 1 ) // Get the first date of that month.
.plusMonths( 1 ) // Get first of next month for Half-Open query.
Details
Assuming your column in MySQL is of DATE type…
LocalDate
The LocalDate class represents a date-only value without time-of-day and without time zone.
Time zone
A time zone is crucial in determining a date. For any given moment, the date varies around the globe by zone. For example, a few minutes after midnight in Paris France is a new day while still “yesterday” in Montréal Québec.
Specify a proper time zone name in the format of continent/region, such as America/Montreal, Africa/Casablanca, or Pacific/Auckland. Never use the 3-4 letter abbreviation such as EST or IST as they are not true time zones, not standardized, and not even unique(!).
ZoneId z = ZoneId.of( "America/Montreal" );
LocalDate today = LocalDate.now( z );
YearMonth
The YearMonth class represents an entire month. Getting the current month requires a time zone as discussed above. Around the beginning/ending of the month, the current moment could be “next” month in Auckland New Zealand while still “previous” month in Kolkata India.
YearMonth currentMonth = YearMonth.now( z ) ;
Get the first date of the month.
LocalDate start = currentMonth.atDayOfMonth( 1 ) ;
Half-Open
Generally best to use the Half-Open [) approach to defining a span of time, where the beginning is inclusive while the ending is exclusive. So defining a month means starting with the first date of the month and running up to, but not including, the first date of the following month.
LocalDate stop = start.plusMonths( 1 ) ;
Query
Do not use the BETWEEN command in SQL as it is fully closed [], both beginning and ending being inclusive. Half-Open uses >= & < logic.
SELECT when FROM tbl
WHERE when >= start
AND when < stop
;
it's also useful
#Query("from PogWorkTime p where p.codePto = :codePto and month(p.dateApply) = :month and year(p.dateApply) = :year")
Iterable<PtoExceptWorkTime> findByCodePtoAndDateApply_MonthAndDateApply_Year(#Param("codePto") String codePto,#Param("month") int month, #Param("year") int year);
I have a custom list that I'm trying to restrict data entry for valid day of week and time.
My current column validation works for day of week being Monday, Wednesday or Friday. It looks like this:
=CHOOSE(WEEKDAY([Requested date for approval]),FALSE,TRUE,FALSE,TRUE,FALSE,TRUE,FALSE)
I'm trying to figure out the syntax to add that it also has to be between 8 am and 12:00 pm on those days.
Any help would be greatly appreciated.
You would use an AND statement to include a second criteria
=AND(CHOOSE(WEEKDAY([Requested date for approval]),FALSE,TRUE,FALSE,TRUE,FALSE,TRUE,FALSE),
AND(
[Requested date for approval]-INT([Requested date for approval])*24 >= 8,
[Requested date for approval]-INT([Requested date for approval])*24 <= 24
)
)
I confess, I've never heard of the CHOOSE function, but the time calculation is based on the information at Microsoft
Convert times
To convert hours from the standard time format to a decimal number, use the INT
function.
Column1 Formula Description (possible result)
10:35 AM =([Column1]-INT([Column1]))*24 Number of hours since 12:00 AM (10.583333)
12:15 PM =([Column1]-INT([Column1]))*24 Number of hours since 12:00 AM (12.25)
EDIT
To calculate the day of the week, you can use the TEXT function to return the day of the week (i.e. Monday)
=TEXT(WEEKDAY([ColumnName]), "dddd")
It won't be pretty, but you can use a series of AND logical operators
=AND(
TEXT(WEEKDAY([Requested date for approval]), "dddd") = "Monday",
AND(
TEXT(WEEKDAY([Requested date for approval]), "dddd") = "Wednesday",
AND(
TEXT(WEEKDAY([Requested date for approval]), "dddd") = "Friday",
AND(
[Requested date for approval]-INT([Requested date for approval])*24 >= 8,
[Requested date for approval]-INT([Requested date for approval])*24 <= 24
)
)
)
)
Posting Working Solution
=IF(
AND(
CHOOSE(
WEEKDAY([Requested date for approval]),FALSE,TRUE,FALSE,TRUE,FALSE,TRUE,FALSE
),
([Requested date for approval]-INT([Requested date for approval]))*24>=8,
([Requested date for approval]-INT([Requested date for approval]))*24<=12
),
TRUE
)
How do I subtract 30 days from a date in PowerBuilder?
I have the following code that returns today's date in a parameter, but I need today - 30 days:
dw_1.setitem(1, "begin_datetime",
datetime(today(), Now()))
you're probably looking for the RelativeDate function. Unfortunately, it takes a Date and not a DateTime as a parameter, but we can get around that:
dw_1.setItem(1, "begin_datetime", DateTime( RelativeDate( today(), -30), Now() )