查询实体框架,包括尚未提交的记录

itc*_*chi 5 c# entity-framework .net-3.5

是否可以查询上下文,对实体进行更改,然后重新查询上下文并获取包含已更改实体的结果集而不执行保存更改?(我不想保存更改,因为我可能想要回滚)

我知道我可以通过使用GetObjectStateEntries来获取更改,但我对整个数据集感兴趣...而不仅仅是更改的实体.

我认为实现这一目标的唯一方法是使用savechanges,但是将事务管理器包装在所有内容中,这样如果我的条件不满足,我就可以进行回滚.还是我更容易错过一些东西?

Mer*_*itt 3

为什么不将现有实体集与要添加的实体集合并?我对此进行了测试,它似乎有效 - 虽然它不考虑删除,但您应该能够明白这个想法:

// get the entities that have been inserted or modified
var projects = myObjectContext.ObjectStateManager.GetObjectStateEntries(
     EntityState.Added | EntityState.Modified).Where(
          x => x.Entity is Project).Select( x=> (Project) x.Entity);

// get existing entities, exclude those that are being modified
var projects2 = myObjectContext.Projects.Where(
     BuildDoesntContainExpression<Project, int>(z => z.ProjectId, 
          projects.Select(x => x.ProjectId)));

// Union the 2 sets
var projects3 = projects.Union(projects2);
Run Code Online (Sandbox Code Playgroud)

BuildDoesntContainExpression:你不能使用 contains,因此你不能使用 EF 进行相反的操作,由于某种原因,所以使用这个方法:

    private static Expression<Func<TElement, bool>> BuildDoesntContainExpression<TElement, TValue>(    
        Expression<Func<TElement, TValue>> valueSelector, IEnumerable<TValue> values)
    {    
        if (null == valueSelector) 
        { 
            throw new ArgumentNullException("valueSelector"); 
        }

        if (null == values) 
        { 
            throw new ArgumentNullException("values"); 
        }    

        ParameterExpression p = valueSelector.Parameters.Single();    

        // p => valueSelector(p) == values[0] || valueSelector(p) == ... 
        if (!values.Any())    
        {        
            return e => false;    
        }    

        var equals = values.Select(
            value => (Expression)Expression.NotEqual(valueSelector.Body, Expression.Constant(value, typeof(TValue))));    

        var body = equals.Aggregate<Expression>((accumulate, equal) => Expression.And(accumulate, equal));    

        return Expression.Lambda<Func<TElement, bool>>(body, p);
     }
Run Code Online (Sandbox Code Playgroud)