c# - 自动比较属性

Question

我想获取为匹配对象而更改的所有属性的名称。我有这些(简化的)类:

public enum PersonType { Student, Professor, Employee }

class Person {
    public string Name { get; set; }
    public PersonType Type { get; set; }
}

class Student : Person {
     public string MatriculationNumber { get; set; }
}

class Subject {
     public string Name { get; set; }
     public int WeeklyHours { get; set; }
}

class Professor : Person {
    public List<Subject> Subjects { get; set; }
}

现在我想获取属性值不同的对象:

List<Person> oldPersonList = ...
List<Person> newPersonList = ...
List<Difference> = GetDifferences(oldPersonList, newPersonList);

public List<Difference> GetDifferences(List<Person> oldP, List<Person> newP) {
     //how to check the properties without casting and checking 
     //for each type and individual property??
     //can this be done with Reflection even in Lists??
}

最后，我想要一个这样的差异列表:

class Difference {
    public List<string> ChangedProperties { get; set; }
    public Person NewPerson { get; set; }
    public Person OldPerson { get; set; }
}

ChangedProperties 应包含已更改属性的名称。

Answer 1

我花了很长时间尝试使用类型化委托(delegate)编写一个更快的基于反射的解决方案。但最终我放弃了，转而使用Marc Gravell's。 Fast-Member library获得比普通反射更高的性能。

代码:

internal class PropertyComparer
{    
    public static IEnumerable<Difference<T>> GetDifferences<T>(PropertyComparer pc,
                                                               IEnumerable<T> oldPersons,
                                                               IEnumerable<T> newPersons)
        where T : Person
    {
        Dictionary<string, T> newPersonMap = newPersons.ToDictionary(p => p.Name, p => p);
        foreach (T op in oldPersons)
        {
            // match items from the two lists by the 'Name' property
            if (newPersonMap.ContainsKey(op.Name))
            {
                T np = newPersonMap[op.Name];
                Difference<T> diff = pc.SearchDifferences(op, np);
                if (diff != null)
                {
                    yield return diff;
                }
            }
        }
    }

    private Difference<T> SearchDifferences<T>(T obj1, T obj2)
    {
        CacheObject(obj1);
        CacheObject(obj2);
        return SimpleSearch(obj1, obj2);
    }

    private Difference<T> SimpleSearch<T>(T obj1, T obj2)
    {
        Difference<T> diff = new Difference<T>
                                {
                                    ChangedProperties = new List<string>(),
                                    OldPerson = obj1,
                                    NewPerson = obj2
                                };
        ObjectAccessor obj1Getter = ObjectAccessor.Create(obj1);
        ObjectAccessor obj2Getter = ObjectAccessor.Create(obj2);
        var propertyList = _propertyCache[obj1.GetType()];
        // find the common properties if types differ
        if (obj1.GetType() != obj2.GetType())
        {
            propertyList = propertyList.Intersect(_propertyCache[obj2.GetType()]).ToList();
        }
        foreach (string propName in propertyList)
        {
            // fetch the property value via the ObjectAccessor
            if (!obj1Getter[propName].Equals(obj2Getter[propName]))
            {
                diff.ChangedProperties.Add(propName);
            }
        }
        return diff.ChangedProperties.Count > 0 ? diff : null;
    }

    // cache for the expensive reflections calls
    private Dictionary<Type, List<string>> _propertyCache = new Dictionary<Type, List<string>>();
    private void CacheObject<T>(T obj)
    {
        if (!_propertyCache.ContainsKey(obj.GetType()))
        {
            _propertyCache[obj.GetType()] = new List<string>();
            _propertyCache[obj.GetType()].AddRange(obj.GetType().GetProperties().Select(pi => pi.Name));
        }
    }
}

用法:

PropertyComparer pc = new PropertyComparer();
var diffs = PropertyComparer.GetDifferences(pc, oldPersonList, newPersonList).ToList();

性能:

我非常有偏见的测量表明，这种方法比 Json 转换快 4-6 倍，比普通反射快约 9 倍。但公平地说，您可能会大大加快其他解决方案的速度。

限制:

目前，我的解决方案不会对嵌套列表进行递归，例如，它不会比较单个 Subject 项目 - 它只会检测到主题列表不同，但不会检测内容或位置。但是，在需要时添加此功能应该不会太难。最困难的部分可能是决定如何在 Difference 类中表示这些差异。