SQL DISTINCT/GROUP BY不能正常工作

Question

忍耐與我在這裏，我對SQL很陌生。我有下面的查詢寫出來，但DISTINCT基本上被我的WHERE語句否定。這是由於表格中的日期後跟一個時間戳，所以如果訂單項放置時間有差異，它會生成一個新行。

什麼是最好的方式讓它完全忽略日期後的所有內容，以便不會因時間戳而生成新行？從我發現的情況來看，我認爲使用GROUP BY可以更好地實現，但接縫總是會產生錯誤（很可能我不知道將聚合函數放在哪裏）。

任何幫助將不勝感激。謝謝！

Select DISTINCT 
so.num AS Ref 
, so.shiptoname AS Recipient_Full_Name 
, so.shiptoaddress AS Address_1 
, so.shiptocity AS City 
, stateconst.name AS State 
, so.shiptozip AS Zip 
, so.billtoname AS Buyer_Name 
, qbclass.name AS Class 
, soitem.datescheduledfulfillment AS Fulfillment_Date 
From SO 
JOIN stateconst 
ON so.shiptostateid=stateconst.id 
JOIN qbclass 
ON so.qbclassid=qbclass.id 
JOIN soitem 
ON so.id=soitem.soid 
WHERE DATESCHEDULEDFULFILLMENT LIKE '2016-05-10%' 

Answer 1

使用CAST()到datetime數據類型更改爲date類型：

Select DISTINCT 
so.num AS Ref 
, so.shiptoname AS Recipient_Full_Name 
, so.shiptoaddress AS Address_1 
, so.shiptocity AS City 
, stateconst.name AS State 
, so.shiptozip AS Zip 
, so.billtoname AS Buyer_Name 
, qbclass.name AS Class 
, CAST(soitem.datescheduledfulfillment as date) AS Fulfillment_Date 
From SO 
JOIN stateconst 
ON so.shiptostateid=stateconst.id 
JOIN qbclass 
ON so.qbclassid=qbclass.id 
JOIN soitem 
ON so.id=soitem.soid 
WHERE CAST(soitem.datescheduledfulfillment as date) = '2016-05-10' 

Answer 2

你真的需要在查詢輸出的日期？因爲離開它會解決您的「問題」：

Select DISTINCT 
so.num AS Ref 
, so.shiptoname AS Recipient_Full_Name 
, so.shiptoaddress AS Address_1 
, so.shiptocity AS City 
, stateconst.name AS State 
, so.shiptozip AS Zip 
, so.billtoname AS Buyer_Name 
, qbclass.name AS Class 

From SO 
JOIN stateconst 
ON so.shiptostateid=stateconst.id 
JOIN qbclass 
ON so.qbclassid=qbclass.id 
JOIN soitem 
ON so.id=soitem.soid 
WHERE DATESCHEDULEDFULFILLMENT LIKE '2016-05-10%'