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)
Related
I want to get month value using week no. I have week numbers stored in a table with year value
How to query database to get month value using that week value.
I am using hiveQL.
Eample:
201801(week) ------->201801(month)
201804(week) ------->201801(month)
Suppose week has full 7 days. Extract week number, multiply 7, add to 'year-01-01' and extract month from result date, lpad with 0 to get '01' from 1 and concatenate with year:
with data_example as(
select stack(3,'201801','201804','201805') as yr_wk
)
select yr_wk, concat(substr(yr_wk, 1,4),
lpad(month(date_add(date(concat(substr(yr_wk, 1,4),'-01-01')),int(substr(yr_wk, 5,2))*7)),2,0)
) yr_mth
from data_example
Result:
yr_wk yr_mth
201801 201801
201804 201801
201805 201802
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 am trying to use the so called DateDiff function to subtract the End Date from the Start Date and obtain the numbers of days apart.For example:
10/11/1995 - 7/11/1995 = 3 (extract the 'dd' from DD/MM/YYYY format)
As date values are double with the integer part counting for a day, you can use this simple expression:
[Due Date]-[Start Date]
or, for integer days only:
Fix([Due Date]-[Start Date])
That said, you should a query for tasks like this.
Let's say that I have a range of SQL tables that are named name_YYYY_WW where YYYY = year and WW = week number. If I call upon a function that guides a user defined date to the right table.
If the date entered is "20110101":
SELECT EXTRACT (WEEK FROM DATE '20110101') returns 52 and
SELECT EXTRACT (YEAR FROM DATE '20110101') returns 2011.
While is nothing wrong with these results I want "20110101" to either point to table name_2010_52 or name_2011_01, not name_2011_52 as it does now when I concanate the results to form the query for the table.
Any elegant solutions to this problem?
The function to_char() will allow you to format a date or timestamp to output correct the iso week and iso year.
SELECT to_char('2011-01-01'::date, 'IYYY_IW') as iso_year_week;
will produce:
iso_year_week
---------------
2010_52
(1 row)
You could use a CASE:
WITH sub(field) AS (
SELECT CAST('20110101' AS date) -- just to test
)
SELECT
CASE
WHEN EXTRACT (WEEK FROM field ) > 1 AND EXTRACT (MONTH FROM field) = 1 AND EXTRACT (DAY FROM field) < 3 THEN 1
ELSE
EXTRACT (WEEK FROM field)
END
FROM
sub;
let's say that I have a SmallDateTime column in my table. How to update hours in each row in T-SQL ?
Get the target value's hour part.
Find the difference between the hour you want and the found hour.
Add the difference of hours to the target value.
The script:
UPDATE atable
SET datetimevalue = DATEADD(hour, #hour - DATEPART(hour, datetimevalue),
datetimevalue)
WHERE ...
UPDATE YourTable SET YourDateColumn = DateAdd(hh, 1, YourDateColumn)
will add 1 hour to the time value in YourDateColumn in every single row.
See this page for more info.
UPDATE YourTableName
SET YourSmallDateTimeColumn = DATEADD(HH, 1, YourSmallDateTimeColumn)
Will add any number of hours to your column. So if you get rid of the time component first:
SELECT CAST(CONVERT(char(8), YourSmallDateTimeColumn, 112) AS smalldatetime)
and add hour component to it later it should work.
From here
To add or subtract hours to a datetime
or smalldatetime value, you will use
the DATEADD date function. The
DATEADD date function returns a new
datetime value based on adding an
interval to the specified date. The
syntax of the DATEADD date function is
as follows:
DATEADD ( datepart , number, date )
datepart is the parameter that
specifies on which part of the date to
return a new value. For hours, you
can use either HOUR or HH. number is
the value used to increment datepart.
date is an expression that returns a
datetime or smalldatetime value, or a
character string in a date format.
Here's an example on how to use the
DATEADD date function to increase or
decrease a datetime value by a certain
number of hours:
SELECT DATEADD(HOUR, -12, GETDATE())AS [TwelveHoursAgo]
SELECT DATEADD(HH, 6, GETDATE()) AS [SixHoursLater]