如何从日期 PostgreSQL 中提取月份中的星期几

Question

I would like to extract from a DATE type the number of the week within the month (NOT in the year).

Example : today's date is 07/08/2018. I do NOT want 32 as a result (week in the year) but 2 , since today is the second week of August.

I would expect something like

SELECT EXTRACT($func FROM current_date);

with result: 2

Where the first week of the month lasts from the first day of the month until the first Sunday (which may last from 1 to 7 days for the first week, eg for this month the first week lasts from Wednesday 1/8/2018 to Sunday 5/8/2018, 5 days).

Adding the calendar of this month to give more context to users:

Answer 1

Try extracting the day, and the dividing by 7:

SELECT
    1 + FLOOR((EXTRACT(DAY FROM date_col) - 1) / 7) AS week_of_month
FROM your_table;

Demo

Answer 2

Another option would be to use TO_CHAR() , like this:

SELECT TO_CHAR( CURRENT_DATE, 'W' )::integer

However this returns 1 , too, as it counts weeks FROM the 1st of the month... (Thus not from a Monday or a Sunday.)

Answer 3

Your week of the month is business data. Usually, business data belongs in tables.

Data in tables is easier to audit than complicated date arithmetic. (Witness all the incorrect date arithmetic in the answers here.) It's simpler to implement arbitrary logic with data in tables. It's easier to understand and troubleshoot, too.

create table my_calendar (
  cal_date date primary key, 
  cal_year integer not null 
    check (cal_year = extract (year from cal_date)),
  cal_month integer not null 
    check (cal_month = extract (month from cal_date)),
  cal_week integer not null 
    check (cal_week between 1 and 5)
);

I included the year and month (cal_year and cal_month), because that's the next thing everyone seems to need.

Some starter data.

insert into my_calendar values
('2018-08-01',  2018, 8, 1),
('2018-08-02',  2018, 8, 1),
('2018-08-03',  2018, 8, 1),
('2018-08-04',  2018, 8, 1),
('2018-08-05',  2018, 8, 1),
('2018-08-06',  2018, 8, 2),
('2018-08-07',  2018, 8, 2),
('2018-08-08',  2018, 8, 2),
('2018-08-09',  2018, 8, 2),
('2018-08-10',  2018, 8, 2),
('2018-08-11',  2018, 8, 2),
('2018-08-12',  2018, 8, 2),
('2018-08-13',  2018, 8, 3),
('2018-08-14',  2018, 8, 3),
('2018-08-15',  2018, 8, 3);

And a query to verify the logic.

select cal_date
from my_calendar
where cal_year = 2018 
  and cal_month = 8 
  and cal_week = 1
order by cal_date;

cal_date
date
--
2018-08-01
2018-08-02
2018-08-03
2018-08-04
2018-08-05

Carefully control privileges in this kind of table. You probably want everybody to be able to read from it, but almost nobody should be able to write to it.

Answer 4

Old question, and found many poorly formatted & explained or wrong answers. For whoever comes next, here's the right answer:

SELECT CEILING((DATE_PART( 'day', CURRENT_DATE ) - DATE_PART( 'dow', CURRENT_DATE )) / 7) +1;

Simply: Find the prior Sunday, divide by 7 and (optionally) add 1.

Counterintuitively it'll calc. a negative value for the first fractional (sometimes full) week, but since it's fractional it'll round up (CEILING) to a zero. The final "+1" bring it to integers 1 through 5 instead of -0 through 4.

See dbfiddle.uk: https://dbfiddle.uk/?rdbms=postgres_9.4&fiddle=f20ec2a530f0b33c673f669d07717e4e

WITH _dt (dt, i ) AS (SELECT * FROM 
     ( SELECT '2022-01-01'::DATE + i AS dt , i 
        FROM  (SELECT generate_series( 0, 365, 1 ) i ) i ( i) 
      ) dt
) 
SELECT DISTINCT
    dt
    ,                                     DATE_PART( 'month', dt ) AS month
    ,                                     DATE_PART( 'dow' , dt )  AS day_of_week
    ,CEILING((DATE_PART( 'day', dt.dt ) - DATE_PART( 'dow', dt.dt )) / 7) +1  AS week_of_month
    ,ROUND(((DATE_PART( 'day', dt.dt ) - DATE_PART( 'dow', dt.dt )) / 7)::NUMERIC, 2 ) AS raw_wom
    ,DATE_PART( 'year', dt ) AS year
    FROM _dt dt   
    ORDER BY year, month, week_of_month, day_of_week
    

Answer 5

I've tested this & it appears to work, weeks 0 - 5:

SELECT CEILING((DATE_PART( 'day', CURRENT_DATE ) - DATE_PART( 'dow', CURRENT_DATE )) / 7);

Simply: Find the prior Sunday, divide by 7.

However for months where the first week starts on Sunday (a full week), an adjustment needs to be made so its ranking behavior (0 - 5) is the same as other months (see flag "is_1st_week_7") . **

Note that in raw form it calculates (somewhat counterintuitively) a negative value for the first week, but it'll round up (CEILING) to a zero.

** (my prior reply had this functional bug)

at dbfiddle.uk: https://dbfiddle.uk/?rdbms=postgres_10&fiddle=e70251943588251fb1883fa6fa44a2dd

WITH _dt (dt, i ) AS (SELECT *
FROM  ( SELECT '2023-01-01'::DATE + i AS dt , i FROM  (SELECT generate_series( 0, 365, 1 ) i ) i ( i) ) dt
) 
SELECT
*
,        ROUND((((dom) - dow) / 7)::NUMERIC, 2 )  - is_1st_week_7 AS raw_wom /** raw week of month **/
,CEILING(ROUND((((dom) - dow) / 7)::NUMERIC, 2 )) - is_1st_week_7 AS wom /** week of month **/
FROM
(
SELECT DISTINCT
 -- full first weeks start the month off with 1 instead of 0, so need to decrement by 1 (above)
 (DATE_PART( 'dow'  , (dt.dt - ((DATE_PART( 'day', dt.dt )::INT)-1)))=0)::INT AS is_1st_week_7
,dt.dt                               /* the date    */
,DATE_PART( 'month', dt )    AS mo   /* month       */  
,DATE_PART( 'day'  , dt.dt ) AS dom  /* day of month*/  
,DATE_PART( 'dow'  , dt )    AS dow  /* day of week */  
,DATE_PART( 'year' , dt )    AS yr   /* year        */ 
FROM _dt dt
) dt
ORDER BY dt 
;