Down below is the SQL that creates the "Orders" table. It includes every version of an Order. In the code below, I used it to retrieve Order S20026. I returns its 7 versions.

SELECT DISTINCT

sha.[Bill-to Name]AS [CustomerNameBillTo]

,sha.[Bill-to Customer No_]AS [CustomerNumberBillTo]

,sha.[Sell-to Customer No_]AS [CustomerNumberSellTo]

,sha.[Sell-to Customer Name]AS [CustomerNameSellTo]

,sha.No_AS [SalesOrderNumber]

,sha.[Version No_]AS [VersionNumber]

,sla.[No_]AS [ItemNumber]

,sha.[Order Date]AS [OrderDate]

,sha.[Requested Delivery Date]AS [RequestedDeliveryDate]

,DATEDIFF(DAY, sha.[Order Date], sha.[Requested Delivery Date]) AS [LeadTimeReqDeliveryDate]

,sha.[Promised Delivery Date] AS [PromisedDeliveryDate]

,DATEDIFF(DAY, sha.[Order Date], sha.[Promised Delivery Date]) AS [LeadTimePromisedDeliveryDate]

,sla.QuantityAS [QuantityOrdered]

,sha.[Posting Date]AS [PostingDate]

,CONVERT(INT, REPLACE(sha.[Shipping Time], CHAR(2), '')) AS [ShippingTime]

,CASE

WHEN YEAR(sha.[Posting Date]) <> '1753'

THEN DATEADD(DAY, CONVERT(INT, REPLACE(sha.[Shipping Time], CHAR(2), '')), sha.[Posting Date])

ELSE NULL

ENDAS [ActualDeliveryDate]

,NULLAS [LeadTimeActualDeliveryDate]

,sla.[Quantity Shipped]AS [ShippedQuantity]

FROM NAV.dbo.[Sales Header Archive] sha

LEFT OUTER JOIN NAV.dbo.[Sales Line Archive] sla

ON sha.[No_] = sla.[Document No_]

AND sha.[Version No_] = sla.[Version No_]

WHERE sla.[Type] = 2

AND sha.No_ = 'S20026'

I noticed that if I run your code against the Orders table the results are accurate.

When I replace every instance of the Orders table with the sql above (which I think is twice in your code) then the results are different. It returns an additional version