SQL數據范圍內的按日期分組

Question

在sql中查詢如下表的最佳方法是什么？

我需要確定PackListGrp變量（ PICKONLY或JITPICK由） expectedshipdate ， deliverymethod對於給定的OrderID 。

當expectedshipdate是在2天內為每個訂單項則PackListGrp用於將JITPICK 。

當expectedshipdate DeliveryMethod日期> 2天時，“提貨DeliveryMethod將為“ PickOnly而“ PSIEST JITPICK 。

基本上，我只是在試圖組合由Expectedshipdate確定的訂單項。

謝謝

OrderID ExpectedShipDate    Sku     DeliveryMethod  PackListGrp
66064   13-Sep-18         22-81506  Pick            PickOnly
66064   13-Sep-18         878-85487 Pick            PickOnly
66064   13-Sep-18         5-4878-   Pick            PickOnly
66064   13-Sep-18         3-020-02  Pick            PickOnly
66064   13-Sep-18         7-209-23  Pick            PickOnly
66064   13-Sep-18         7-206-05  Pick            PickOnly
66064   13-Sep-18         7-305-08  Pick            PickOnly
66064   13-Sep-18         7-567-20  Pick            PickOnly
66064   18-Sep-18         6-104-24  Pick            JITPICK
66064   19-Sep-18         TCC_KS    PSIEST          JITPICK
66064   17-Sep-18         TCF-DK    PSIEST          JITPICK

Answer 1

您是否只需要一個case表達式？

select t.*,
       (case when datediff(day, t. expectedshipdate, getdate()) <= 2
             then 'JITPICK'
             else 'PICKONLY'
        end) as packlistgrp
from t;

Answer 2

如果我理解正確，那么您想將PackListGroup用作計算字段。

如果正確，此代碼將為您提供幫助

SELECT
    CASE
        WHEN ExpectedShipDate BETWEEN DATEADD(DAY,-2,GETDATE()) AND GETDATE() THEN 'JITPICK'
        WHEN ExpectedShipDate > DATEADD(DAY,-2,GETDATE()) THEN 'PickOnly'
        ELSE 'JITPICK' END AS PackListGrp
FROM
    yourTable