Question

我正在尝试在Linq查询中引用内联表值函数（ITVF）：

var results = await (
    from v in _context.Vehicles
    from r in _context.UnitRepairStatus(v.VehicleNumber) <-- ITVF reference
    orderby v.VehicleNumber
    select new FooViewModel { 
            ID = v.ID, 
            VehicleNumber = v.VehicleNumber,
            InRepair = Convert.ToBoolean(r.InRepair) <-- ITFV field
        }
    ).ToListAsync();

查询运行时，会生成错误：

InvalidOperationException：第二个操作在此上下文上开始在上一个操作完成之前。任何实例成员都不是保证是线程安全的。

提及代码：

System.Linq.AsyncEnumerable.Aggregate_（IAsyncEnumerable源，TAccumulate种子， Func累加器，Func resultSelector，CancellationToken取消cancelToken） MyApplication.Controllers.VehiclesController.Foo（）在 VehiclesController.cs
                var results = await (

如果删除ITFV引用，查询将按预期工作

var results = await (
    from v in _context.Vehicles
    orderby v.VehicleNumber
    select new FooViewModel { 
            ID = v.ID, 
            VehicleNumber = v.VehicleNumber,
            InRepair = False <-- dummy value
        }
    ).ToListAsync();

为什么添加ITVF参考时会发生这种情况？我该如何解决？

代码

UnitRepairStatus ITVF：

CREATE FUNCTION dbo.UnitRepairStatus(
  @unit_number varchar(18)
)
RETURNS TABLE
AS

RETURN

  SELECT  h.InRepair
  -- connects to a second database on same server
  --  shouldn't be an issue, but mentioning it in case it might be
  FROM    Schema2..Unit u
  INNER JOIN Schema2..History h on u.ID = h.UnitID
  WHERE   u.UnitNumber = @unit_number

UnitRepairStatus型号：

public class UnitRepairStatus
{
    public string UnitNumber { get; set; }
    public int? InRepair { get; set; }
}

MyDatabaseDbContext DbContext：

public class MyDatabaseDbContext : DbContext
{

    public MyDatabaseDbContext(DbContextOptions<MyDatabaseDbContext> options) : base(options) {}
...

    protected override void OnModelCreating(ModelBuilder modelBuilder)
    {
        ...
        modelBuilder.Query<UnitRepairStatus>();
    }

    public IQueryable<UnitRepairStatus> UnitRepairStatus(string unitNumber) =>
        Query<UnitRepairStatus>().FromSql($"SELECT * FROM UnitRepairStatus({unitNumber})");

}

FooViewModel ViewModel：

public class FooViewModel
{
    public int ID { get; set; }
    public string VehicleNumber { get; set; }
    public bool InRepair { get; set; }
}

VehiclesController构造函数：

public VehiclesController(
    ILogger<VehiclesController> logger, 
    MyDatabaseDbContext context
)
{
    _logger = logger;
    _context = context;
}

VehiclesController Foo方法：

public async Task<IActionResult> Foo()
{

    List<FooViewModel> model = null;

    try
    {
        var results = await (  <-- line referenced in error message
            from v in _context.Vehicles
            from r in _context.UnitRepairStatus(v.VehicleNumber)
            orderby v.VehicleNumber
            select new FooViewModel { 
                    ID = v.ID, 
                    VehicleNumber = v.VehicleNumber,
                    InRepair = Convert.ToBoolean(r.InRepair)
                }
            ).ToListAsync();

    }
    catch (Exception e)
    {
        _logger.LogError(e.Message);
        throw;
    }

    return View(model);

}

参考：

Use a Inline Table-Valued Functions with Linq and Entity Framework

Answer 1

对不起，我的错。上一个问题的答案中的技术适用于调用具有常数/变量参数的ITVF，但不适用于与您的情况相关的子查询（以及我的错误示例）。

解决方案是删除ITVF参数并将结果扩展为也包括该列（有效地将其转换为无参数视图）：

CREATE FUNCTION dbo.UnitRepairStatus()
RETURNS TABLE
AS
RETURN
  SELECT u.UnitNumber, h.InRepair
  FROM    Schema2.Unit u
  INNER JOIN Schema2.History h on u.ID = h.UnitID

还要从上下文方法中删除参数：

public IQueryable<UnitRepairStatus> UnitRepairStatus() =>
    Query<UnitRepairStatus>().FromSql("SELECT * FROM UnitRepairStatus()");

，然后将LINQ查询更改为使用join：

var results = await (
    from v in _context.Vehicles
    join r in _context.UnitRepairStatus() on v.VehicleNumber equals r.UnitNumber // <---
    orderby v.VehicleNumber
    select new FooViewModel { 
        ID = v.ID, 
        VehicleNumber = v.VehicleNumber,
        InRepair = Convert.ToBoolean(r.InRepair)
    }
).ToListAsync();

现在，它应该转换并执行服务器端，并在客户端成功实现。

原始方法的问题是EF Core默默地将查询执行切换为客户端评估（对此讨厌），然后达到了在同一上下文中执行多个异步操作的保护。

引用ITVF会引发“在此上下文中开始的第二次操作，而之前的操作尚未完成”

代码

1 个答案: