postgresql - Postgres : creating DATE Dimension as per Ralph Kimball

Question

我是 postgres 和 DW 的新手，我必须设计一个 DATE Dimension，如 book 中给出的那样.我在网上看了很多地方，但到目前为止我没有成功，能否解释一下如何填充 'Fiscal Week'、'Fiscal Month'、'Fiscal Half year'

等字段

谢谢

Answer 1

我猜您想要 10 年的日期维度表，其中包含图 2.4 中显示的所有列(基于 Google Books )。 checkin 文档:

• 格式化函数to_char与 Table 9-21用于模板模式
• 提取函数 date_part
• 使用 generate_series(start, stop, step interval) 生成行系列

要获得 10 年内的所有天数，您可以这样写:

SELECT generate_series('2001-01-01'::date, '2010-12-31'::date, '1 day') AS day;

SELECT count(*) FROM generate_series('2001-01-01'::date, '2010-12-31'::date, '1 day');
 count 
-------
  3652
(1 row)

根据图2.5建表如:

DROP TABLE IF EXISTS "Date Dimension";
CREATE TABLE "Date Dimension"
(
    "Date Key" serial,
    "Date" date,
    "Full Day Description" text,
    "Day Of Week" text,
    "Calendar Month" text,
    "Calendar Year" integer,
    "Fiscal Year Month" text,
    "Holiday Indicator" text,
    "Weekday Indicator" text 
);

插入命令:

INSERT INTO "Date Dimension"
    ("Date", "Full Day Description", "Day Of Week", "Calendar Month",
    "Calendar Year", "Fiscal Year Month", "Holiday Indicator",
    "Weekday Indicator")
SELECT
    day,
    rtrim(to_char(day, 'Month')) || to_char(day, ' DD, YYYY'),
    to_char(day, 'Day'),
    rtrim(to_char(day, 'Month')),
    date_part('year', day),
    'F' || to_char(day, 'YYYY-MM'),
    '', --omitting (trivial 'Holiday'/'Non-Holiday, but how to get this ??),
    CASE
        WHEN date_part('isodow', day) IN (6, 7) THEN 'Weekend'
        ELSE 'Weekday'
    END
FROM
    generate_series('2001-01-01'::date, '2010-12-31'::date, '1 day') day;

我希望这能为您提供一些框架和起点。