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EF6不保存更改

entity-framework sql-server c#

GregN · 技术社区 · 6 年前

在下面的代码中,我围绕多个savechanges包装了一个dbContextTransactions。ViewModel中的所有内容都包含正确的数据,但由于某些原因,EF不保存更改。当我调试它时,我没有看到任何异常抛出。有什么特殊的事情我需要做的交易,这将阻止工作?

我已经验证了我的目标确实是正确的数据库,因为我可以通过另一个页面成功地查询它。

    public static Logger logger = LogManager.GetCurrentClassLogger();

    public static Guid AddPlanItem(PlanItemAddViewModel viewModel, IEnumerable<HttpPostedFileBase> images)
    {
        Guid planIdGUID = Guid.NewGuid();

        using (var context = new ApplicationDbContext())
        {
            using (var dbContextTransaction = context.Database.BeginTransaction())
            {
                try
                {
                    PlanItem planItem = context.Plans.Create();

                    planItem.PlanId = planIdGUID;
                    planItem.CreateDate = DateTime.Now;
                    planItem.PlanIdTitle = viewModel.PlanIdTitle;
                    planItem.PlanTitle = viewModel.PlanTitle;
                    planItem.SolutionTitle = viewModel.SolutionTitle;
                    planItem.ActivityCodeId = viewModel.ActivityCodeId;
                    planItem.RegionId = viewModel.RegionId;
                    planItem.OperatingCenterId = viewModel.OperatingCenterId;
                    planItem.DistrictId = viewModel.DistrictId;
                    planItem.ServiceCenterId = viewModel.ServiceCenterId;
                    planItem.PlannerRACFId = viewModel.PlannerRACFId;
                    planItem.SeverityId = viewModel.SeverityId;
                    planItem.Description = viewModel.Description;
                    planItem.PreviousPlan = viewModel.PreviousPlan;

                    context.Plans.Add(planItem);
                    context.SaveChanges();

                    foreach (HttpPostedFileBase image in images)
                    {
                        var imageSubmit = new PlanItemImage
                        {
                            PlanId = planIdGUID,
                            PlanImageId = Guid.NewGuid(),
                            Image = image.ConvertToByte(),
                            ImageTitle = image.FileName
                        };

                        context.Image.Add(imageSubmit);
                        context.SaveChanges();
                    }

                    dbContextTransaction.Commit();
                }
                catch (Exception ex)
                {
                    logger.Error("Error: " + ex.Message);
                    dbContextTransaction.Rollback();
                }
            }
        }

        return planIdGUID;
    }

1 回复 | 直到 6 年前

RaulMonteroc 6 年前

对于我看到的代码,我还不太确定问题可能是什么。即便如此,我们可以采取一种办法来确定问题可能是什么。

首先让我们简化代码。

public static Guid AddPlanItem(PlanItemAddViewModel viewModel, IEnumerable<HttpPostedFileBase> images)
{
    Guid planIdGUID = Guid.NewGuid();

    using (var context = new ApplicationDbContext())
    {
        PlanItem planItem = context.Plans.Create();

        planItem.PlanId = planIdGUID;
        planItem.CreateDate = DateTime.Now;
        planItem.PlanIdTitle = viewModel.PlanIdTitle;
        planItem.PlanTitle = viewModel.PlanTitle;
        planItem.SolutionTitle = viewModel.SolutionTitle;
        planItem.ActivityCodeId = viewModel.ActivityCodeId;
        planItem.RegionId = viewModel.RegionId;
        planItem.OperatingCenterId = viewModel.OperatingCenterId;
        planItem.DistrictId = viewModel.DistrictId;
        planItem.ServiceCenterId = viewModel.ServiceCenterId;
        planItem.PlannerRACFId = viewModel.PlannerRACFId;
        planItem.SeverityId = viewModel.SeverityId;
        planItem.Description = viewModel.Description;
        planItem.PreviousPlan = viewModel.PreviousPlan;

        context.Plans.Add(planItem);
        context.SaveChanges();
    }

    return planIdGUID;
}

通过删除try/catch,我们可以看到任何可能的异常
通过简化保存逻辑,我们可以确定问题所在。如果这个代码一切正常,那么我们就可以包含保存图像的片段以及它的行为。

第二,如果一切正常,我们可以添加下一块,图像,但这一次与保存更改以外的颊。

foreach (HttpPostedFileBase image in images)
{
    var imageSubmit = new PlanItemImage
    {
        PlanId = planIdGUID,
        PlanImageId = Guid.NewGuid(),
        Image = image.ConvertToByte(),
        ImageTitle = image.FileName
    };

    context.Image.Add(imageSubmit);    
}

context.SaveChanges();

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