请按照附图所示的方式帮助您获取数据
突出显示的行实际上在“项目编号”字段
上分组提前致谢
答案 0 :(得分:1)
这是一种方式:
SELECT
it.[Item Number]
, Values1 = STUFF(CAST((
SELECT [text()] = ', ' + CAST(it2.TotalItem AS varchar) + ' x ' + it2.ItemName
FROM dbo.Items it2
WHERE it2.[Item Number] = it.[Item Number]
FOR XML PATH(''), TYPE) AS VARCHAR(100)), 1, 2, '')
FROM (
SELECT DISTINCT [Item Number]
FROM dbo.Items
) it
答案 1 :(得分:0)
以下应该工作:
SELECT distinct ItemNumber,SUBSTRING(
(SELECT ';' + TotalItem+' x '+ItemName
FROM YourTable
where ItemNumber=c.ItemNumber
ORDER BY ItemNumber
FOR XML PATH('')),2,200000 ) AS NewDescription from YourTable c
GO
答案 2 :(得分:0)
您可以使用 STRING_AGG
SELECT [Item Number], STRING_AGG (CONCAT(TotalItem,'x',ItemName), ';') as tags
FROM dbo.TableName
GROUP BY [Item Number];
答案 3 :(得分:0)
SELECT t2.ItemNumber ,
STUFF(( SELECT ',' + CAST(TotalItem AS VARCHAR) + ' x ' + ItemName
FROM @MyTable t1
WHERE t1.ItemNumber = t2.ItemNumber
FOR
XML PATH('')
), 1, 1, '') AS NewDescription
FROM @MyTable t2
GROUP BY t2.ItemNumber