我正在寻找插入实体框架的最快方法。

我之所以问这个问题,是因为您有一个活动的TransactionScope,并且插入量很大(4000+)。它可能会持续超过10分钟(事务的默认超时),这将导致事务不完整。


当前回答

〔2019更新〕EF Core 3.1

如上所述,在EF Core中禁用AutoDetectChangesEnabled非常有效:插入时间除以100(从几分钟到几秒,10k条记录具有交叉表关系)

更新的代码为:

context.ChangeTracker.AutoDetectChangesEnabled = false;
foreach (IRecord record in records) {
    //Add records to your database        
}
context.ChangeTracker.DetectChanges();
context.SaveChanges();
context.ChangeTracker.AutoDetectChangesEnabled = true; //do not forget to re-enable

其他回答

因为这里从未提到过,我想在这里重新推荐EFCore.BulkExtensions

context.BulkInsert(entitiesList);                 context.BulkInsertAsync(entitiesList);
context.BulkUpdate(entitiesList);                 context.BulkUpdateAsync(entitiesList);
context.BulkDelete(entitiesList);                 context.BulkDeleteAsync(entitiesList);
context.BulkInsertOrUpdate(entitiesList);         context.BulkInsertOrUpdateAsync(entitiesList);         // Upsert
context.BulkInsertOrUpdateOrDelete(entitiesList); context.BulkInsertOrUpdateOrDeleteAsync(entitiesList); // Sync
context.BulkRead(entitiesList);                   context.BulkReadAsync(entitiesList);

秘密是插入到相同的空白暂存表中。插件快速发光。然后在主大表中运行一个单独的插入。然后截断临时表,为下一批做好准备。

ie.

insert into some_staging_table using Entity Framework.

-- Single insert into main table (this could be a tiny stored proc call)
insert into some_main_already_large_table (columns...)
   select (columns...) from some_staging_table
truncate table some_staging_table

您可以使用Bulk包库。大容量插入1.0.0版本用于实体框架>=6.0.0的项目。

更多描述可在此处找到-Bulkoperation源代码

我对上面的@Slauma示例进行了一个通用扩展;

public static class DataExtensions
{
    public static DbContext AddToContext<T>(this DbContext context, object entity, int count, int commitCount, bool recreateContext, Func<DbContext> contextCreator)
    {
        context.Set(typeof(T)).Add((T)entity);

        if (count % commitCount == 0)
        {
            context.SaveChanges();
            if (recreateContext)
            {
                context.Dispose();
                context = contextCreator.Invoke();
                context.Configuration.AutoDetectChangesEnabled = false;
            }
        }
        return context;
    }
}

用法:

public void AddEntities(List<YourEntity> entities)
{
    using (var transactionScope = new TransactionScope())
    {
        DbContext context = new YourContext();
        int count = 0;
        foreach (var entity in entities)
        {
            ++count;
            context = context.AddToContext<TenancyNote>(entity, count, 100, true,
                () => new YourContext());
        }
        context.SaveChanges();
        transactionScope.Complete();
    }
}

[POSTGRESQL的新解决方案]嘿,我知道这是一篇很老的文章,但我最近遇到了类似的问题,但我们使用的是Postgresql。我想使用有效的膨胀剂,但结果很难。我在这个数据库上找不到合适的免费库。我只找到了这个助手:https://bytefish.de/blog/postgresql_bulk_insert/也在Nuget上。我编写了一个小映射器,它以实体框架的方式自动映射财产:

public static PostgreSQLCopyHelper<T> CreateHelper<T>(string schemaName, string tableName)
        {
            var helper = new PostgreSQLCopyHelper<T>("dbo", "\"" + tableName + "\"");
            var properties = typeof(T).GetProperties();
            foreach(var prop in properties)
            {
                var type = prop.PropertyType;
                if (Attribute.IsDefined(prop, typeof(KeyAttribute)) || Attribute.IsDefined(prop, typeof(ForeignKeyAttribute)))
                    continue;
                switch (type)
                {
                    case Type intType when intType == typeof(int) || intType == typeof(int?):
                        {
                            helper = helper.MapInteger("\"" + prop.Name + "\"",  x => (int?)typeof(T).GetProperty(prop.Name).GetValue(x, null));
                            break;
                        }
                    case Type stringType when stringType == typeof(string):
                        {
                            helper = helper.MapText("\"" + prop.Name + "\"", x => (string)typeof(T).GetProperty(prop.Name).GetValue(x, null));
                            break;
                        }
                    case Type dateType when dateType == typeof(DateTime) || dateType == typeof(DateTime?):
                        {
                            helper = helper.MapTimeStamp("\"" + prop.Name + "\"", x => (DateTime?)typeof(T).GetProperty(prop.Name).GetValue(x, null));
                            break;
                        }
                    case Type decimalType when decimalType == typeof(decimal) || decimalType == typeof(decimal?):
                        {
                            helper = helper.MapMoney("\"" + prop.Name + "\"", x => (decimal?)typeof(T).GetProperty(prop.Name).GetValue(x, null));
                            break;
                        }
                    case Type doubleType when doubleType == typeof(double) || doubleType == typeof(double?):
                        {
                            helper = helper.MapDouble("\"" + prop.Name + "\"", x => (double?)typeof(T).GetProperty(prop.Name).GetValue(x, null));
                            break;
                        }
                    case Type floatType when floatType == typeof(float) || floatType == typeof(float?):
                        {
                            helper = helper.MapReal("\"" + prop.Name + "\"", x => (float?)typeof(T).GetProperty(prop.Name).GetValue(x, null));
                            break;
                        }
                    case Type guidType when guidType == typeof(Guid):
                        {
                            helper = helper.MapUUID("\"" + prop.Name + "\"", x => (Guid)typeof(T).GetProperty(prop.Name).GetValue(x, null));
                            break;
                        }
                }
            }
            return helper;
        }

我使用它的方式如下(我有一个名为“承诺”的实体):

var undertakingHelper = BulkMapper.CreateHelper<Model.Undertaking>("dbo", nameof(Model.Undertaking));
undertakingHelper.SaveAll(transaction.UnderlyingTransaction.Connection as Npgsql.NpgsqlConnection, undertakingsToAdd));

我展示了一个事务的示例,但它也可以通过从上下文中检索到的正常连接来完成。undertakingsToAdd是普通实体记录的枚举,我想将其批量插入数据库。

这个解决方案,我经过几个小时的研究和尝试后得到的,正如你所期望的那样,速度更快,最终易于使用和免费!我真的建议你使用这个解决方案,不仅因为上面提到的原因,而且因为它是唯一一个我对Postgresql本身没有问题的解决方案,许多其他解决方案都可以完美地工作,例如SqlServer。