从 Oracle SQL 上的周和年数字获取“一周开始”的日期

Question

I have been working with a query that groups a number of records by weeks. The week numbers are extracted by

to_char(reportdate, 'IW') as "Week",   

after which a GROUP BY clause is used. The "Week" always starts on a Monday and if the 1st January is not a Monday, Week 1 will start from the previous Monday (or so as I've understood this line of code).

So if a table such as the following is obtained (for three example rows)

Week  Year  Sales  Visits
 32   2017    22     55
 33   2017    30     65
 01   2019    55     103

I'd like to add a column with the date such that it is the start of the week. I was thinking of using the grouped query as a sub-query, then simply adding a the column after, but I can't exactly figure out how to map the Week and Year number to such a date.

I've tried this line to obtain the date

trunc(next_day   (trunc(to_date(Yr,'yy'),'yy')  -1,'Mon') + (7*(Wk - 1))    -7)  

But it has not been holding true for each row.

Any ideas to point me in the right direction would be appreciated.

Answer 1

ISO-8601 counts the first week of the year as the first week which has the majority of its days in that year. To have the majority of its days in that year then it must contain at least 4 days of the week within that year and must contain the 4th January.

You can use this to work out the start of the first iso-week as:

• TO_DATE( year, 'YYYY' ) will give 1st January;
• Then you can add 3 days to get to the 4th January; and
• Use TRUNC( date_value, 'IW' ) to get the Monday of the 1st iso-week of the year
• Then just add the correct offset of weeks to go from the 1st to Nth week.

Test Data :

CREATE TABLE test_data ( Week, Year, Sales, Visits ) AS
SELECT '32', 2017, 22,  55 FROM DUAL UNION ALL
SELECT '33', 2017, 30,  65 FROM DUAL UNION ALL
SELECT '01', 2019, 55, 103 FROM DUAL;

Query :

SELECT t.*,
       TRUNC( TO_DATE( year, 'YYYY' ) + 3, 'IW' ) + 7 * ( week - 1 ) AS week_start
FROM   test_data t

Output :

\nWEEK | YEAR | SALES | VISITS | WEEK_START \n:--- |  ---: |  ----: |  -----: |  :--------- \n32 |  2017 |  22 |  55 |  07-AUG-17  \n33 |  2017 |  30 |  65 |  14-AUG-17  \n01 |  2019 |  55 |  103 |  31-DEC-18  \n

db<>fiddle here

Answer 2

According to ISO-8601 the first week is the week having January 4th. I use this function to get date from ISO-Week:

CREATE OR REPLACE FUNCTION ISOWeekDate(WEEK INTEGER, YEAR INTEGER) RETURN DATE DETERMINISTIC IS
    res DATE;
BEGIN
    IF WEEK > 53 OR WEEK < 1 THEN
        RAISE VALUE_ERROR;      
    END IF;
    res := NEXT_DAY(TO_DATE( YEAR || '0104', 'YYYYMMDD' ) - 7, 'MONDAY') + ( WEEK - 1 ) * 7;
    IF TO_CHAR(res, 'fmIYYY') = YEAR THEN
        RETURN res;
    ELSE
        RAISE VALUE_ERROR;
    END IF;
END ISOWeekDate;

Note, the week number without year is not sufficient, because ISO-Year can be different to actual year. See these examples:

SELECT to_char(DATE '2019-12-31', 'IYYY-"W"IW') from dual;

TO_CHAR(DATE'2019-12-31','IYYY-"W"IW')
--------------------------------------
2020-W01                              


SELECT ISOWeekDate(1, 2019) from dual;

ISOWEEKDATE(1,2019)
----------------------------
2018-12-31 00:00:00         

Of course, when your input week/year values are reliable then you can simply use SELECT NEXT_DAY(TO_DATE( yr|| '0104', 'YYYYMMDD' ) - 7, 'MONDAY') + ( wk - 1 ) * 7 AS week_start