我有2个表有多对多的关系如下
部门表
╔═══════════╦════════════════╗ ║ deptID ║ DeptName ║ ╠═══════════╬════════════════╣ ║ 1 ║ abc ║ ║ 2 ║ def ║ ║ 3 ║ ghi ║ ╚═══════════╩════════════════╝
另一个是 标准表
╔════════╦════════════════════════╗ ║ id ║ criteria ║ ╠════════╬════════════════════════╣ ║ 1 ║ number of Employees ║ ║ 2 ║ status ║ ║ 3 ║ nature of products ║ ║ . ║ --- ║ ║ . ║ --- ║ ║ . ║ --- ║ ║ . ║ --- ║ ║ . ║ --- ║ ║ . ║ --- ║ ║ . ║ --- ║ ╚════════╩════════════════════════╝
在单独的表中以多对多的关系动态增加，为每个部门提供一个值，如下所示
╔════════╦══════╦═════════════════════╗ ║ DeptID ║ CrID ║ value ║ ╠════════╬══════╬═════════════════════╣ ║ 1 ║ 1 ║ 20 ║ ║ 1 ║ 2 ║ currently active ║ ║ 1 ║ 3 ║ mechanical ║ ║ 1 ║ . ║ ..... ║ ║ 1 ║ . ║ ..... ║ ║ 1 ║ . ║ ..... ║ ║ 2 ║ 1 ║ 40 ║ ║ 2 ║ 2 ║ paused for transfer ║ ║ 2 ║ 3 ║ software ║ ║ 2 ║ . ║ ..... ║ ║ 2 ║ . ║ ..... ║ ║ 2 ║ . ║ ..... ║ ║ 3 ║ 1 ║ 50 ║ ║ 3 ║ 2 ║ heavy duty ║ ║ 3 ║ 3 ║ support ║ ║ 3 ║ . ║ ..... ║ ║ 3 ║ . ║ ..... ║ ║ 3 ║ . ║ ..... ║ ║ ║ ║ ║ ║ . ║ . ║ ..... ║ ║ . ║ . ║ ..... ║ ║ . ║ . ║ ..... ║ ║ ║ ║ ║ ╚════════╩══════╩═════════════════════╝
1）如何获得此结果
╔═══════╦═════════════╦═════════════════════╦════════════╦═════╦═════╗ ║ name ║ number of ║ status ║ nature of ║ ... ║ ... ║ ║ ║ Employees ║ ║ products ║ ║ ║ ╠═══════╬═════════════╬═════════════════════╬════════════╬═════╬═════╣ ║ abc ║ 20 ║ currently active ║ mechanical ║ ... ║ ... ║ ║ def ║ 40 ║ paused for transfer ║ software ║ ... ║ ... ║ ║ ghi ║ 50 ║ heavy duty ║ support ║ ... ║ ... ║ ║ . ║ .. ║ ... ║ ... ... ║ ║ ║ ║ . ║ .. ║ ... ║ ... ... ║ ║ ║ ║ . ║ .. ║ ... ║ ... ... ║ ║ ║ ╚═══════╩═════════════╩═════════════════════╩════════════╩═════╩═════╝
2）如何获得一个表如下
╔══════╦═══════════════════════════════════════════════════════════════════════════════════╗ ║ name ║ notes ║ ╠══════╬═══════════════════════════════════════════════════════════════════════════════════╣ ║ abc ║ number of Employees 20 , status currently active , nature of products mechanical ║ ║ def ║ number of Employees 40 , status paused for transfer , nature of products software ║ ║ ghi ║ number of Employees 50 , status heavy duty, nature of products support ║ ║ ║ ║ ║ ... ║ .................................................. ║ ║ ... ║ .................................................. ║ ║ ... ║ .................................................. ║ ║ ║ ║ ╚══════╩═══════════════════════════════════════════════════════════════════════════════════╝
答案 0 :(得分：2)
（1）
你需要这个结果吗？这将需要使用动态SQL，因为标准的数量未知或未修复，因此结果中的列数不固定
（2）
SELECT d.DeptName,
notes = stuff(n.notes, 1, 1, '')
FROM Departments d
OUTER APPLY
(
SELECT ',' + c.Criteria + ' ' + dc.value
FROM DeptCriteria dc
INNER JOIN Criteria c ON dc.CrID = c.id
WHERE dc.DeptID = d.deptID
ORDER BY c.id
FOR XML PATH('')
) n (notes)