如何在Redshift中創建日期表？

tl; dr：我想在Redshift中生成一個日期表，以便更容易地生成報告。不需要大型表已經在redshift，需要上傳一個csv文件。如何在Redshift中創建日期表？

長版本：我正在編寫一份報告，我必須平均每週創建新項目。日期範圍可能會持續數月或更長時間，所以可能會有5個星期一，但只有4個星期日，這可能會使數學有點棘手。另外，我無法保證每天有單個項目的實例，特別是一旦用戶開始分割數據。其中，這正在絆倒BI工具。

解決此問題的最佳方法很可能是日期表。但是，日期表的大多數教程都使用了Redshift無法提供或不完全支持的SQL命令（我在看着你，generate_series）。

有沒有一種簡單的方法在Redshift中生成日期表？

我嘗試使用的代碼（在此基礎上也 - 不工作的建議：http://elliot.land/post/building-a-date-dimension-table-in-redshift）

CREATE TABLE facts.dates (
    "date_id"    INTEGER      NOT NULL PRIMARY KEY, 

    -- DATE 
    "full_date"   DATE      NOT NULL, 

    -- YEAR 
    "year_number"   SMALLINT     NOT NULL, 
    "year_week_number"  SMALLINT     NOT NULL, 
    "year_day_number"  SMALLINT     NOT NULL, 

    -- QUARTER 
    "qtr_number"   SMALLINT     NOT NULL, 

    -- MONTH 
    "month_number"   SMALLINT     NOT NULL, 
    "month_name"   CHAR(9)      NOT NULL, 
    "month_day_number"  SMALLINT     NOT NULL, 

    -- WEEK 
    "week_day_number"  SMALLINT     NOT NULL, 

    -- DAY 
    "day_name"    CHAR(9)      NOT NULL, 
    "day_is_weekday"  SMALLINT     NOT NULL, 
    "day_is_last_of_month" SMALLINT     NOT NULL 
) DISTSTYLE ALL SORTKEY (date_id) 
; 


INSERT INTO facts.dates 
(
    "date_id" 
    ,"full_date" 
    ,"year_number" 
    ,"year_week_number" 
    ,"year_day_number" 

    -- QUARTER 
    ,"qtr_number" 

    -- MONTH 
    ,"month_number" 
    ,"month_name" 
    ,"month_day_number" 

    -- WEEK 
    ,"week_day_number" 

    -- DAY 
    ,"day_name" 
    ,"day_is_weekday" 
    ,"day_is_last_of_month" 
) 
    SELECT 
    cast(seq + 1 AS INTEGER)          AS date_id, 

    -- DATE 
    datum               AS full_date, 

    -- YEAR 
    cast(extract(YEAR FROM datum) AS SMALLINT)     AS year_number, 
    cast(extract(WEEK FROM datum) AS SMALLINT)     AS year_week_number, 
    cast(extract(DOY FROM datum) AS SMALLINT)      AS year_day_number, 

    -- QUARTER 
    cast(to_char(datum, 'Q') AS SMALLINT)       AS qtr_number, 

    -- MONTH 
    cast(extract(MONTH FROM datum) AS SMALLINT)     AS month_number, 
    to_char(datum, 'Month')          AS month_name, 
    cast(extract(DAY FROM datum) AS SMALLINT)      AS month_day_number, 

    -- WEEK 
    cast(to_char(datum, 'D') AS SMALLINT)       AS week_day_number, 

    -- DAY 
    to_char(datum, 'Day')           AS day_name, 
    CASE WHEN to_char(datum, 'D') IN ('1', '7') 
     THEN 0 
    ELSE 1 END             AS day_is_weekday, 
    CASE WHEN 
     extract(DAY FROM (datum + (1 - extract(DAY FROM datum)) :: INTEGER + 
         INTERVAL '1' MONTH) :: DATE - 
         INTERVAL '1' DAY) = extract(DAY FROM datum) 
     THEN 1 
    ELSE 0 END             AS day_is_last_of_month 
    FROM 
    -- Generate days for 81 years starting from 2000. 
    (
     SELECT 
     '2000-01-01' :: DATE + generate_series AS datum, 
     generate_series      AS seq 
     FROM generate_series(0,81 * 365 + 20,1) 
    ) DQ 
    ORDER BY 1;

會拋出這個錯誤

[Amazon](500310) Invalid operation: Specified types or functions (one per INFO message) not supported on Redshift tables.; 
1 statement failed.

......因爲，我假設INSERT和generate_series不允許在Redshift中的同一命令中

來源

2017-11-10 Phillip

正如你已經發現，'generate_series（）'不能與實際的數據，因爲它僅執行領導節點上使用。你的方法生成一個數字表，然後加入它的效果很好。或者，在Excel中創建源文件並僅導入結果。像這樣的日期表非常適合報告。您可能想要添加的其他內容：公共假期標誌，季度標誌的最後一天，年份標誌的最後一天（適用於按期間最後一個日期分組的報告）。 –

我喜歡那些額外的列。謝謝約翰！ – Phillip

作爲一個工作karound，您可以在本地計算機上運行Postgres實例，在那裏運行代碼，導出爲CSV，然後僅在Redshift中運行CREATE TABLE部分並從CSV加載數據。由於這是一次性操作，因此可以這麼做，這就是我實際爲新的Redshift部署所做的事情。

來源

2017-11-10 20:54:29 AlexYes

非常好的主意，但我想出了一個方法來做到這一點，而無需上傳csv。不幸的是，會採取一些複製粘貼魔法。如果您有任何改進，我在下面發佈我的解決方案。 – Phillip

在問這個問題時，我明白了。哎呀。

我從一個「事實」模式開始。

CREATE SCHEMA facts;

運行下面開始數表：

create table facts.numbers 
(
    number int PRIMARY KEY 
) 
;

使用此生成你的電話號碼清單。我用一百萬上手

SELECT ',(' || generate_series(0,1000000,1) || ')' 
;

然後數從結果在下面的查詢複製粘貼，值之後：

INSERT INTO facts.numbers 
VALUES 
(0) 
,(1) 
,(2) 
,(3) 
,(4) 
,(5) 
,(6) 
,(7) 
,(8) 
,(9) 
-- etc

^確保從禁止複製刪除前導逗號數字

的粘貼名單一旦你有一個數字表，那麼你就可以生成一個日期表（再次，從艾略特土地http://elliot.land/post/building-a-date-dimension-table-in-redshift偷碼）：

CREATE TABLE facts.dates (
    "date_id"    INTEGER      NOT NULL PRIMARY KEY, 

    -- DATE 
    "full_date"   DATE      NOT NULL, 

    -- YEAR 
    "year_number"   SMALLINT     NOT NULL, 
    "year_week_number"  SMALLINT     NOT NULL, 
    "year_day_number"  SMALLINT     NOT NULL, 

    -- QUARTER 
    "qtr_number"   SMALLINT     NOT NULL, 

    -- MONTH 
    "month_number"   SMALLINT     NOT NULL, 
    "month_name"   CHAR(9)      NOT NULL, 
    "month_day_number"  SMALLINT     NOT NULL, 

    -- WEEK 
    "week_day_number"  SMALLINT     NOT NULL, 

    -- DAY 
    "day_name"    CHAR(9)      NOT NULL, 
    "day_is_weekday"  SMALLINT     NOT NULL, 
    "day_is_last_of_month" SMALLINT     NOT NULL 
) DISTSTYLE ALL SORTKEY (date_id) 
; 


INSERT INTO facts.dates 
(
    "date_id" 
    ,"full_date" 
    ,"year_number" 
    ,"year_week_number" 
    ,"year_day_number" 

    -- QUARTER 
    ,"qtr_number" 

    -- MONTH 
    ,"month_number" 
    ,"month_name" 
    ,"month_day_number" 

    -- WEEK 
    ,"week_day_number" 

    -- DAY 
    ,"day_name" 
    ,"day_is_weekday" 
    ,"day_is_last_of_month" 
) 
    SELECT 
    cast(seq + 1 AS INTEGER)          AS date_id, 

    -- DATE 
    datum               AS full_date, 

    -- YEAR 
    cast(extract(YEAR FROM datum) AS SMALLINT)     AS year_number, 
    cast(extract(WEEK FROM datum) AS SMALLINT)     AS year_week_number, 
    cast(extract(DOY FROM datum) AS SMALLINT)      AS year_day_number, 

    -- QUARTER 
    cast(to_char(datum, 'Q') AS SMALLINT)       AS qtr_number, 

    -- MONTH 
    cast(extract(MONTH FROM datum) AS SMALLINT)     AS month_number, 
    to_char(datum, 'Month')          AS month_name, 
    cast(extract(DAY FROM datum) AS SMALLINT)      AS month_day_number, 

    -- WEEK 
    cast(to_char(datum, 'D') AS SMALLINT)       AS week_day_number, 

    -- DAY 
    to_char(datum, 'Day')           AS day_name, 
    CASE WHEN to_char(datum, 'D') IN ('1', '7') 
     THEN 0 
    ELSE 1 END             AS day_is_weekday, 
    CASE WHEN 
     extract(DAY FROM (datum + (1 - extract(DAY FROM datum)) :: INTEGER + 
         INTERVAL '1' MONTH) :: DATE - 
         INTERVAL '1' DAY) = extract(DAY FROM datum) 
     THEN 1 
    ELSE 0 END             AS day_is_last_of_month 
    FROM 
    -- Generate days for 81 years starting from 2000. 
    (
     SELECT 
     '2000-01-01' :: DATE + number AS datum, 
     number      AS seq 
     FROM facts.numbers 
     WHERE number between 0 and 81 * 365 + 20 
    ) DQ 
    ORDER BY 1;

^務必在結束日期範圍設置號碼，你需要

來源

2017-11-10 21:12:56 Phillip

如何在Redshift中創建日期表？

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