如何在psql中获得当前月份的周日数？

如何在postgresql中获取给定date的星期天总数

Answer 1

给定date的星期日总数只能是0或1。

但是如果你想在一个给定的date范围内的星期天的数量，那么你最好的赌注是一个日历表。 要知道今年2月份有多less个星期天，我只是想

select count(*) from calendar where cal_date between '2011-02-01' and '2011-02-28' and day_of_week = 'Sun'; 

要么

 select count(*) from calendar where year_of_date = 2011 and month_of_year = 2 and day_of_week = 'Sun'; 

这是一个基本的日历表，你可以开始。 我还包括一个PostgreSQL函数来填充日历表。 我没有在8.3中testing过，但是我确定我没有使用8.3不支持的function。

请注意，“dow”部分假定你的日子是英文的。 但是，您可以轻松地编辑这些部分以匹配任何语言。 （我想，但是我可能错在“轻松”）。

 -- Table: calendar -- DROP TABLE calendar; CREATE TABLE calendar ( cal_date date NOT NULL, year_of_date integer NOT NULL, month_of_year integer NOT NULL, day_of_month integer NOT NULL, day_of_week character(3) NOT NULL, CONSTRAINT calendar_pkey PRIMARY KEY (cal_date), CONSTRAINT calendar_check CHECK (year_of_date::double precision = date_part('year'::text, cal_date)), CONSTRAINT calendar_check1 CHECK (month_of_year::double precision = date_part('month'::text, cal_date)), CONSTRAINT calendar_check2 CHECK (day_of_month::double precision = date_part('day'::text, cal_date)), CONSTRAINT calendar_check3 CHECK (day_of_week::text = CASE WHEN date_part('dow'::text, cal_date) = 0::double precision THEN 'Sun'::text WHEN date_part('dow'::text, cal_date) = 1::double precision THEN 'Mon'::text WHEN date_part('dow'::text, cal_date) = 2::double precision THEN 'Tue'::text WHEN date_part('dow'::text, cal_date) = 3::double precision THEN 'Wed'::text WHEN date_part('dow'::text, cal_date) = 4::double precision THEN 'Thu'::text WHEN date_part('dow'::text, cal_date) = 5::double precision THEN 'Fri'::text WHEN date_part('dow'::text, cal_date) = 6::double precision THEN 'Sat'::text ELSE NULL::text END) ) WITH ( OIDS=FALSE ); ALTER TABLE calendar OWNER TO postgres; -- Index: calendar_day_of_month -- DROP INDEX calendar_day_of_month; CREATE INDEX calendar_day_of_month ON calendar USING btree (day_of_month); -- Index: calendar_day_of_week -- DROP INDEX calendar_day_of_week; CREATE INDEX calendar_day_of_week ON calendar USING btree (day_of_week); -- Index: calendar_month_of_year -- DROP INDEX calendar_month_of_year; CREATE INDEX calendar_month_of_year ON calendar USING btree (month_of_year); -- Index: calendar_year_of_date -- DROP INDEX calendar_year_of_date; CREATE INDEX calendar_year_of_date ON calendar USING btree (year_of_date); 

还有一个基本的函数来填充表格。 我也没有在8.3中testing过这个。

 -- Function: insert_range_into_calendar(date, date) -- DROP FUNCTION insert_range_into_calendar(date, date); CREATE OR REPLACE FUNCTION insert_range_into_calendar(from_date date, to_date date) RETURNS void AS $BODY$ DECLARE this_date date := from_date; BEGIN while (this_date <= to_date) LOOP INSERT INTO calendar (cal_date, year_of_date, month_of_year, day_of_month, day_of_week) VALUES (this_date, extract(year from this_date), extract(month from this_date), extract(day from this_date), case when extract(dow from this_date) = 0 then 'Sun' when extract(dow from this_date) = 1 then 'Mon' when extract(dow from this_date) = 2 then 'Tue' when extract(dow from this_date) = 3 then 'Wed' when extract(dow from this_date) = 4 then 'Thu' when extract(dow from this_date) = 5 then 'Fri' when extract(dow from this_date) = 6 then 'Sat' end); this_date = this_date + interval '1 day'; end loop; END; $BODY$ LANGUAGE plpgsql VOLATILE COST 100; 

Answer 2

你需要提取物：

 SELECT EXTRACT(DOW FROM DATE '2011-02-16') = 0; -- 0 is Sunday 

这可能会导致真实或错误，这是一个星期天或不是。 我不知道你的意思是“总数”，因为那总是0（date不是星期天）或1（给定的数据是星期天）。

编辑：这样的东西？

 SELECT COUNT(*) FROM generate_series(timestamp '2011-01-01', '2011-03-01', '1 day') AS g(mydate) WHERE EXTRACT(DOW FROM mydate) = 0;