27

我的 Asp.net mvc Web 应用程序中有以下 Action 方法:-

[HttpPost]
[ValidateAntiForgeryToken]
public ActionResult Create(SDJoin sdj, FormCollection formValues)
{
    Try
    {
        //code goes here
        repository.InsertOrUpdateSD(sdj.StorageDevice, User.Identity.Name, assetid);
        repository.Save();
    }
    catch (Exception ex)
    {
        //code goes here
    }
    PopulateViewBagData();
    return View(sdj);
}

它调用以下方法:-

public void InsertOrUpdateSD(TMSStorageDevice sd, string username, long assetid)
{
    var resource = entities.Resources.AsNoTracking().SingleOrDefault(a => a.RESOURCEID == assetid);
    if (sd.TMSStorageDeviceID == default(int))
    {
        // New entity
        int technologyypeID = GetTechnologyTypeID("Storage Device");
        Technology technology = new Technology
        {
            IsDeleted = true,
            IsCompleted = false,
            TypeID = technologyypeID,
            Tag = "SD" + GetTagMaximumeNumber2(technologyypeID).ToString(),
            StartDate = DateTime.Now,
            IT360ID = assetid
        };

        InsertOrUpdateTechnology(technology);
        Save();

        sd.TMSStorageDeviceID = technology.TechnologyID;
        tms.TMSStorageDevices.Add(sd);
    }
}

我的模型类如下:-

public partial class TMSStorageDevice
{
    public int TMSStorageDeviceID { get; set; }
    public string Name { get; set; }
    public Nullable<decimal> size { get; set; }
    public int RackID { get; set; }
    public string CustomerName { get; set; }
    public string Comment { get; set; }
    public byte[] timestamp { get; set; }

    public virtual Technology Technology { get; set; }
    public virtual TMSRack TMSRack { get; set; }
}

但是如果我调用 Create 操作方法,我会得到以下异常:-

System.Data.Entity.Validation.DbEntityValidationException was caught
  HResult=-2146232032
  Message=Validation failed for one or more entities. See 'EntityValidationErrors' property for more details.
  Source=EntityFramework
  StackTrace:
       at System.Data.Entity.Internal.InternalContext.SaveChanges()
       at System.Data.Entity.Internal.LazyInternalContext.SaveChanges()
       at System.Data.Entity.DbContext.SaveChanges()
       at TMS.Models.Repository.Save() in c:\Users\Administrator\Documents\Visual Studio 2012\Projects\TMS\TMS\Models\Repository.cs:line 1926
       at TMS.Controllers.StorageDeviceController.Create(SDJoin sdj, FormCollection formValues) in c:\Users\Administrator\Documents\Visual Studio 2012\Projects\TMS\TMS\Controllers\StorageDeviceController.cs:line 160
  InnerException:

任何人都可以建议有什么问题,因为我仔细检查了我的代码并且每件事都应该正常工作?谢谢

4

3 回答 3

130

You haven't shown the Save() method but if you can add code like this to it you'll get an exception that contains all the details you're looking for

try
{
    _context.SaveChanges();
}
catch (System.Data.Entity.Validation.DbEntityValidationException dbEx)
{
    Exception raise = dbEx;
    foreach (var validationErrors in dbEx.EntityValidationErrors)
    {
        foreach (var validationError in validationErrors.ValidationErrors)
        {
            string message = string.Format("{0}:{1}", 
                validationErrors.Entry.Entity.ToString(),
                validationError.ErrorMessage);
            // raise a new exception nesting
            // the current instance as InnerException
            raise = new InvalidOperationException(message, raise);
        }
    }
    throw raise;
}
于 2013-10-22T14:59:05.433 回答
7

我知道这个问题现在很老了,但我希望有人也觉得这很有用。

  1. 最有可能(实际上,根据您的评论判断)错误是您的变量未通过模型验证。
  2. 查看您看不到的错误(无需更改代码)的另一种方法是使用 QuickWatch 窗口查看异常。问题是您可以从异常工具提示打开的“查看详细信息”窗口不显示 EntityValidationErrors。如果没有 catch 块,您将无法在 QuickWatch 中查看任何异常变量。幸运的是,QuickWatch 有 PseudoVariables。 因此,使用“$exception”伪变量,您可以轻松查看 QuickWatch 窗口中当前未处理的异常。(请看下面的截图) 异常详细信息窗口带有 $exception 伪变量的 QuickWatch 窗口
于 2015-10-21T16:39:05.887 回答
1

虽然这是一个旧帖子,但这种方法可以更有成效!

学分

尝试这样的事情

        try
        {
            pcontext.SaveChanges();
        }
       catch (System.Data.Entity.Infrastructure.DbUpdateConcurrencyException ex)
        {             
             Console.WriteLine(ex.InnerException);
        }
        catch (System.Data.Entity.Core.EntityCommandCompilationException ex)
        {
          Console.WriteLine(ex.InnerException);
        }
        catch (System.Data.Entity.Core.UpdateException ex)
        {
         Console.WriteLine(ex.InnerException);
        }

        catch (System.Data.Entity.Infrastructure.DbUpdateException ex) //DbContext
        {
            Console.WriteLine(ex.InnerException);
        }

        catch (Exception ex)
        {
            Console.WriteLine(ex.InnerException);
            throw;
        }
于 2017-04-20T12:39:51.713 回答