我正在寻找在 postgres 数据库中创建一个日期表。样本数据预计如下所示：
date key = 00001
calendar_date= 1/1/2015
week_num= 1
month_num= 1
month_name= Jan
quarter_num= 1
calendar_year= 2015
iso_dayofweek= 4
dayofweek_name= Thursday
是否有可以帮助我创建日期 gregorian ISO-8601 表的函数或 SQL？
如果可能的话，我希望自动生成它。在这个方向上的任何帮助将不胜感激。
见下面的例子
SELECT mydate calendar_date
,EXTRACT(WEEK FROM mydate) week_num
,EXTRACT(month FROM mydate) month_num
,to_char(mydate,'Mon') month_name
,EXTRACT(Quarter FROM mydate) quarter_num
,EXTRACT(year FROM mydate) calendar_year
,EXTRACT(DOW FROM mydate) iso_dayofweek
,to_char(mydate, 'day') dayofweek_name
FROM (
SELECT now()::DATE mydate
) t
结果：
calendar_date week_num month_num month_name quarter_num calendar_year iso_dayofweek dayofweek_name
------------- -------- --------- ---------- ----------- ------------- ------------- --------------
2015/04/24 17 4 Apr 2 2015 5 friday
您可以使用generate_series()获取一年中的所有日期，例如：2015
select generate_series(0,364) + date'1/1/2015'
这将产生日期 from ，并在给定的示例中1/1/2015 - 31/12/2015使用此选择而不是SELECT now()::DATE
如果您想为年份创建表，2015则可以使用以下查询
CREATE TABLE mycal_2015 AS
SELECT row_number() OVER () date_key
,mydate calendar_date
,EXTRACT(WEEK FROM mydate) week_num
,EXTRACT(month FROM mydate) month_num
,to_char(mydate,'Mon') month_name
,EXTRACT(Quarter FROM mydate) quarter_num
,EXTRACT(year FROM mydate) calendar_year
,EXTRACT(DOW FROM mydate) iso_dayofweek
,to_char(mydate, 'day') dayofweek_name
FROM (
SELECT generate_series(0, 364) + DATE '1/1/2015' mydate
) t
桌子看起来像select * from mycal_2015
date_key calendar_date week_num month_num month_name quarter_num calendar_year iso_dayofweek dayofweek_name
-------- ------------- -------- --------- ---------- ----------- ------------- ------------- --------------
1 2015/01/01 1 1 Jan 1 2015 4 thursday
2 2015/01/02 1 1 Jan 1 2015 5 friday
3 2015/01/03 1 1 Jan 1 2015 6 saturday
4 2015/01/04 1 1 Jan 1 2015 0 sunday
5 2015/01/05 2 1 Jan 1 2015 1 monday
6 2015/01/06 2 1 Jan 1 2015 2 tuesday
...
.
.
.
364 2015/12/30 53 12 Dec 4 2015 3 wednesday
365 2015/12/31 53 12 Dec 4 2015 4 thursday
PostgreSQL：提取函数
PostgreSQL 提取函数从日期中提取部分
句法 ：extract( unit from date )
date 是要从中提取日期部分的日期、时间戳、时间或间隔值。
unit 是时间间隔的单位类型，如日、月、分、小时等
它可以是以下之一：
unit description
--------------- -----------------------------------------------------------------------------------------------------------------------------
century Uses the Gregorian calendar where the first century starts at '0001-01-01 00:00:00 AD'
day Day of the month (1 to 31)
decade Year divided by 10
dow Day of the week (0=Sunday, 1=Monday, 2=Tuesday, ... 6=Saturday)
doy Day of the year (1=first day of year, 365/366=last day of the year, depending if it is a leap year)
epoch Number of seconds since '1970-01-01 00:00:00 UTC', if date value. Number of seconds in an interval, if interval value
hour Hour (0 to 23)
isodow Day of the week (1=Monday, 2=Tuesday, 3=Wednesday, ... 7=Sunday)
isoyear ISO 8601 year value (where the year begins on the Monday of the week that contains January 4th)
microseconds Seconds (and fractional seconds) multiplied by 1,000,000
millennium Millennium value
milliseconds Seconds (and fractional seconds) multiplied by 1,000
minute Minute (0 to 59)
month Number for the month (1 to 12), if date value. Number of months (0 to 11), if interval value
quarter Quarter (1 to 4)
second Seconds (and fractional seconds)
timezone Time zone offset from UTC, expressed in seconds
timezone_hour Hour portion of the time zone offset from UTC
timezone_minute Minute portion of the time zone offset from UTC
week Number of the week of the year based on ISO 8601 (where the year begins on the Monday of the week that contains January 4th)
year Year as 4-digits
注意：提取功能适用于 PostgreSQL 8.4 及以上版本