SELECT COUNT(DISTINCT I.Id) AS [correctItemCount], FROM [Item] I INNER JOIN [Family] F ON F.[Id] = I.[FamilyId] INNER JOIN [Classification] CL ON CL.[Id] = F.[ClassificationId] LEFT OUTER JOIN [FamilyText] FT ON FT.[Id] = F.[Id] AND FT.[LocaleId] = '1003' --CROSS JOIN [Category] PC ***changed it to leftjoin below INNER JOIN [xCategoryItem] xCI ON xCI.[ItemId] = I.[Id] left join [Category] PC on xCI.[L] BETWEEN PC.[L] AND PC.[R] INNER JOIN [xProduct] xP ON xP.[ItemCode] = I.[Code] AND xP.CanOrderIfActive = 1 INNER JOIN [Category] C ON C.[Id] = xCI.[CategoryId] LEFT OUTER JOIN [CategoryText] CT ON CT.[Id] = C.[Id] AND CT.[LocaleId] = '1003'
Network and collaborate with thousands of CTOs, CISOs, and IT Pros rooting for you and your success.
”The time we save is the biggest benefit of E-E to our team. What could take multiple guys 2 hours or more each to find is accessed in around 15 minutes on Experts Exchange.