【发布时间】：2019-03-14 20:04:34
【问题描述】：
我对 SQL 还很陌生。我试图使用两个查询并将它们连接在一起以获得我想要的结果：
select
A.[Article:Link]
,A.[ArtNo]
,A.[GenericArticle:Link]
,A.[Vehicle:Link]
,A.[LinkingTargetNo]
,Q1.[Name]
FROM [HellaPim].[dbo].[TecDoc.ArticleLinkage.Linkages] A WITH (NOLOCK)
LEFT OUTER JOIN
(SELECT
base.[TecDocNo],
base.[VehicleType],
base.[TypeNo],
T3.[Manufacturer:Link],
T24.[Name],
T1.[Designation],
T1.[LongDesignation],
T2.[DescriptionNo],
base.[Model:Link],
T4.[Designation],
base.[SortNo],
base.[ConstructionYearFrom],
base.[ConstructionYearTo],
base.[KW],
base.[PS],
base.[CcmTaxation],
base.[CcmTechnical],
base.[CapacityLitres],
base.[NumCylinders],
base.[NumDoors],
base.[FuelTankCapacity],
base.[MainVoltage],
base.[ABS],
base.[ASR],
base.[EngineType:Link],
T6.[Description <4>],
base.[FuelMixtureFormation:Link],
T8.[Description <4>],
base.[DriveType:Link],
T10.[Description <4>],
base.[BrakeType:Link],
T12.[Description <4>],
base.[BrakeSystem:Link],
T14.[Description <4>],
base.[NumValves],
base.[FuelType:Link],
T16.[Description <4>],
base.[CatalystConverterType:Link],
T18.[Description <4>],
base.[TransmissionType:Link],
T20.[Description <4>],
base.[BodyType:Link],
T22.[Description <4>]
FROM [HellaPim].[dbo].[TecDoc.LinkingTargets.Types] base WITH (NOLOCK)
LEFT OUTER JOIN [HellaPim].[dbo].[TecDoc.LinkingTargets.Models] T3 WITH (NOLOCK)
ON T3.[ModelNo] = [base].[Model:Link]
LEFT OUTER JOIN [HellaPim].[dbo].[TecDoc.GeneralData.Manufacturers] T23 WITH (NOLOCK)
ON T23.[ManufacturerNo] = [T3].[Manufacturer:Link]
LEFT OUTER JOIN [HellaPim].[dbo].[TecDoc.GeneralData.Manufacturers <TecDoc.GeneralData.Countries>] T24 WITH (NOLOCK)
ON T24.[:Id] = [T23].[:Id]
AND T24.[:TecDoc.GeneralData.Countries_Id] = 'A98E18B0-1401-4104-9EF9-827497E6407F'
LEFT OUTER JOIN [HellaPim].[dbo].[TecDoc.LinkingTargets.Types <TecDoc.GeneralData.Countries>] T1 WITH (NOLOCK)
ON T1.[:Id] = [base].[:Id]
AND T1.[:TecDoc.GeneralData.Countries_Id] = 'D3ECF38D-A5D3-4A46-BED1-24853E25DFC8'
LEFT OUTER JOIN [HellaPim].[dbo].[TecDoc.GeneralData.CouLngDescriptions] T2 WITH (NOLOCK)
ON T2.[DescriptionNo] = [base].[Description:Link]
LEFT OUTER JOIN [HellaPim].[dbo].[TecDoc.LinkingTargets.Models <TecDoc.GeneralData.Countries>] T4 WITH (NOLOCK)
ON T4.[:Id] = [T3].[:Id]
AND T4.[:TecDoc.GeneralData.Countries_Id] = 'D3ECF38D-A5D3-4A46-BED1-24853E25DFC8'
LEFT OUTER JOIN [HellaPim].[dbo].[TecDoc.GeneralData.KeyValues] T5 WITH (NOLOCK)
ON T5.[KeyValueNo] = [base].[EngineType:Link]
LEFT OUTER JOIN [HellaPim].[dbo].[TecDoc.GeneralData.LngDescriptions] T6 WITH (NOLOCK)
ON T6.[DescriptionNo] = [T5].[Name:Link]
LEFT OUTER JOIN [HellaPim].[dbo].[TecDoc.GeneralData.KeyValues] T7 WITH (NOLOCK)
ON T7.[KeyValueNo] = [base].[FuelMixtureFormation:Link]
LEFT OUTER JOIN [HellaPim].[dbo].[TecDoc.GeneralData.LngDescriptions] T8 WITH (NOLOCK)
ON T8.[DescriptionNo] = [T7].[Name:Link]
LEFT OUTER JOIN [HellaPim].[dbo].[TecDoc.GeneralData.KeyValues] T9 WITH (NOLOCK)
ON T9.[KeyValueNo] = [base].[DriveType:Link]
LEFT OUTER JOIN [HellaPim].[dbo].[TecDoc.GeneralData.LngDescriptions] T10 WITH (NOLOCK)
ON T10.[DescriptionNo] = [T9].[Name:Link]
LEFT OUTER JOIN [HellaPim].[dbo].[TecDoc.GeneralData.KeyValues] T11 WITH (NOLOCK)
ON T11.[KeyValueNo] = [base].[BrakeType:Link]
LEFT OUTER JOIN [HellaPim].[dbo].[TecDoc.GeneralData.LngDescriptions] T12 WITH (NOLOCK)
ON T12.[DescriptionNo] = [T11].[Name:Link]
LEFT OUTER JOIN [HellaPim].[dbo].[TecDoc.GeneralData.KeyValues] T13 WITH (NOLOCK)
ON T13.[KeyValueNo] = [base].[BrakeSystem:Link]
LEFT OUTER JOIN [HellaPim].[dbo].[TecDoc.GeneralData.LngDescriptions] T14 WITH (NOLOCK)
ON T14.[DescriptionNo] = [T13].[Name:Link]
LEFT OUTER JOIN [HellaPim].[dbo].[TecDoc.GeneralData.KeyValues] T15 WITH (NOLOCK)
ON T15.[KeyValueNo] = [base].[FuelType:Link]
LEFT OUTER JOIN [HellaPim].[dbo].[TecDoc.GeneralData.LngDescriptions] T16 WITH (NOLOCK)
ON T16.[DescriptionNo] = [T15].[Name:Link]
LEFT OUTER JOIN [HellaPim].[dbo].[TecDoc.GeneralData.KeyValues] T17 WITH (NOLOCK)
ON T17.[KeyValueNo] = [base].[CatalystConverterType:Link]
LEFT OUTER JOIN [HellaPim].[dbo].[TecDoc.GeneralData.LngDescriptions] T18 WITH (NOLOCK)
ON T18.[DescriptionNo] = [T17].[Name:Link]
LEFT OUTER JOIN [HellaPim].[dbo].[TecDoc.GeneralData.KeyValues] T19 WITH (NOLOCK)
ON T19.[KeyValueNo] = [base].[TransmissionType:Link]
LEFT OUTER JOIN [HellaPim].[dbo].[TecDoc.GeneralData.LngDescriptions] T20 WITH (NOLOCK)
ON T20.[DescriptionNo] = [T19].[Name:Link]
LEFT OUTER JOIN [HellaPim].[dbo].[TecDoc.GeneralData.KeyValues] T21 WITH (NOLOCK)
ON T21.[KeyValueNo] = [base].[BodyType:Link]
LEFT OUTER JOIN [HellaPim].[dbo].[TecDoc.GeneralData.LngDescriptions] T22 WITH (NOLOCK)
ON T22.[DescriptionNo] = [T21].[Name:Link]
) as Q1 on A.[LinkingTargetNo] = Q1.[TypeNo]
我的错误是
消息 8156，第 16 级，状态 1，第 107 行 为“Q1”多次指定了“Designation”列。
这两个查询单独工作得很好。问题是当我将它们结合在一起时。如您所见，我在 [Designation] 中输入了两次，但一次来自 T1，一次来自 T4。
我错过了什么？
【问题讨论】：
-
你不能有两个同名的列吗？要么从选择中删除它，要么将其中一个写为
t1.[Designation] as differentname
标签： sql sql-server tsql