I am using Postgres and SQL Server.
Can you test the data pls.

On Tue, 16 Jul 2024 at 16:45, Adrian Klaver <adrian.kla...@aklaver.com>
wrote:

> On 7/16/24 05:53, Anthony Apollis wrote:
> >  1. The problem is the code below reads only data up until 2024,
> >     although the table has been updated with latest data that contains
> >     2025, for some odd reason it is not pulling in or showing data when
> >     filtering for 2025 or even 2024 and later, which should contain all
> >     the latest data.
> >  2.   SQL: "SET DATEFIRST 7;-- Holding Period Query
>
> I don't recognize SET DATEFIRST 7 as Postgres command.
>
> What database are you running this on?
>
> Have not gone through the below fully, but this:
>
> "FY" IS NOT NULL AND "FY" >= 'FY24'
>
> and this
>
> WEEK_NUMBER."Date" < CURRENT_DATE
>
> would seem to be at odds if you want to reach into 2025.
>
> >
> >     SELECT
> >
> >  3.
> >
> >     -- Holding Period Query
> >     SELECT
> >          WEEK_NUMBER."Week Number",
> >          WM_GH."Calendar day",
> >          WM_GH."Month/Week",
> >          WM_GH."Day name",
> >          WM_GH."Company",
> >          WM_GH."Material Code",
> >          WM_GH."Metric",
> >          WM_GH."Metric Value"
> >     FROM
> >          (
> >              SELECT
> >                  "Calendar day",
> >                  "Month/Week",
> >                  "Day name",
> >                  "Company",
> >                  "Material Code",
> >                  "Metric",
> >                  "Metric Value"
> >              FROM
> >
> >     "Prod"."IMETA_Weekly_Metrics_in_Focus_Global_Stock_View_SAP_BW" AS
> WM_GH
> >              WHERE
> >                  WM_GH."Metric Focus" LIKE 'Weekly'
> >                  AND WM_GH."Calendar day" <= (
> >                      SELECT MAX(WEEK_NUMBER."Date") AS "MAX DATE"
> >                      FROM (
> >                          SELECT
> >                              "Date",
> >                              "Week number",
> >                              COUNT("Date") OVER (PARTITION BY "Week
> >     number" ORDER BY "Week number") AS "Number of days",
> >                              COUNT("Date") OVER (PARTITION BY "Week
> >     number" ORDER BY "Date") AS "Day number in weeks"
> >                          FROM (
> >                              SELECT DISTINCT
> >                                  "Date",
> >                                  EXTRACT(WEEK FROM WEEK_NUMBER."Date")
> >     AS "Week number"
> >                              FROM
> >
> >     "dbo"."IMETA_Calendar_Days_Data_Table_Copy10" AS WEEK_NUMBER
> >                              WHERE
> >                                  "FY" IS NOT NULL AND "FY" >= 'FY24'
> >                          ) AS W_MAX
> >                      ) AS WEEK_NUMBER
> >                      WHERE
> >                          WEEK_NUMBER."Date" < CURRENT_DATE
> >                          AND "Number of days" = "Day number in weeks"
> >                  )
> >          ) AS WM_GH
> >     LEFT OUTER JOIN
> >          (
> >              SELECT
> >                  *
> >              FROM (
> >                  SELECT
> >                      "Date",
> >                      "Week number",
> >                      COUNT("Date") OVER (PARTITION BY "Week number"
> >     ORDER BY "Week number") AS "Number of days",
> >                      COUNT("Date") OVER (PARTITION BY "Week number"
> >     ORDER BY "Date") AS "Day number in weeks"
> >                  FROM (
> >                      SELECT DISTINCT
> >                          "Date",
> >                          EXTRACT(WEEK FROM WEEK_NUMBER."Date") AS "Week
> >     number"
> >                      FROM
> >                          "dbo"."IMETA_Calendar_Days_Data_Table_Copy10"
> >     AS WEEK_NUMBER
> >                  ) AS W_MAX
> >              ) AS WEEK_NUMBER
> >              WHERE
> >                  WEEK_NUMBER."Date" < CURRENT_DATE
> >          ) AS WEEK_NUMBER
> >     ON
> >          WM_GH."Calendar day" = WEEK_NUMBER."Date"
> >     ORDER BY
> >          WM_GH."Calendar day" DESC;
> >
>
> --
> Adrian Klaver
> adrian.kla...@aklaver.com
>
>

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