在Linq lambda表达式（GROUP_CONCAT / STRING_AGG）中，XML PATH和Stuff的等价物是什么？

Question

我有这样一张桌子：

EmployeeId  EmployeeName ItemName
4           Ganesh       Key Board
4           Ganesh       Processor
1           Jignesh      Key Board
1           Jignesh      Mouse
1           Jignesh      Processor
3           Rakesh       Key Board
2           Tejas        Key Board
2           Tejas        Mouse
2           Tejas        Processor

我需要查询这仿佛ITEMNAME是相同或不同的employeeid和employeename我们应该有项目，如“”分开。

就像下面给出的那个：

EmployeeId  EmployeeName ItemName
1           Jignesh      Key Board, Mouse, Processor
2           Tejas        Key Board, Mouse, Processor
3           Rakesh       Key Board
4           Ganesh       Key Board, Processor

这是SQL查询：

任何人都可以帮助我将上述SQL查询转换为Lambda表达式吗？

Answer 1

我假设Lambda expression是指Linq语句（例如，对于EF或Linq2Sql）。

显示的FOR XML PATH和STUFF示例是解决Sql Server中缺少GROUP_CONCAT或LISTAGG问题 。 最后在Sql 2017中有STRING_AGG

您根本不需要在LINQ中重现hack - 而只是将感兴趣的员工组的所有行加载到内存中， GroupBy所需的键，然后在选择投影中使用String.Join ：

var result = db.EmployeeItems
      // If you have a filter add the .Where() here ...
      .GroupBy(e => e.EmployeeId)
      .ToList()
      // Because the ToList(), this select projection is not done in the DB
      .Select(eg => new 
       {
          EmployeeId = eg.Key,
          EmployeeName = eg.First().EmployeeName,
          Items = string.Join(",", eg.Select(i => i.ItemName))
       });

employeeItems是Employee和Items之间连接的投影：

var employeeItems = new []
{
   new EmployeeItem{EmployeeId = 1, EmployeeName = "Ganesh", ItemName = "Keyboard"},
   new EmployeeItem{EmployeeId = 1, EmployeeName = "Ganesh", ItemName = "Mouse"},
   new EmployeeItem{EmployeeId = 2, EmployeeName = "John", ItemName = "Keyboard"}
};

结果：

1 Ganesh Keyboard,Mouse
2 John   Keyboard