Question

我查看过本网站上的问题，但没有找到符合我特定问题的内容。

假设我有以下内容：

Product[] store1 = { new Product { Name = "apple", Code = 9, Code1="1" }, 
                   new Product { Name = "orange", Code = 4 } };

Product[] store2 = { new Product { Name = "apple", Code = 9, Code2="2" }, 
                   new Product { Name = "lemon", Code = 12 } };

使用：

public class Product : IEquatable<Product>
{
    public string Name { get; set; }
    public int Code { get; set; }
    public string Code1 { get; set; }
    public string Code2 { get; set; }

    public bool Equals(Product other)
    {

        //Check whether the compared object is null. 
        if (Object.ReferenceEquals(other, null)) return false;

        //Check whether the compared object references the same data. 
        if (Object.ReferenceEquals(this, other)) return true;

        //Check whether the products' properties are equal. 
        return Code.Equals(other.Code) && Name.Equals(other.Name);
    }

    // If Equals() returns true for a pair of objects  
    // then GetHashCode() must return the same value for these objects. 

    public override int GetHashCode()
    {

        //Get hash code for the Name field if it is not null. 
        int hashProductName = Name == null ? 0 : Name.GetHashCode();

        //Get hash code for the Code field. 
        int hashProductCode = Code.GetHashCode();

        //Calculate the hash code for the product. 
        return hashProductName ^ hashProductCode;
    }
}

如何返回单个Enumerable，其中来自store1的数据被匹配的store2中的数据覆盖，并且在非匹配时从store2插入到store1中。基本上我正在寻找TSQL Merge语句的C＃等价物。

运行时的结束时：

foreach (var product in union)
      Console.WriteLine(product.Name + " " + product.Code + " " + product.Code1 + " " + product.Code2);

我想回来：

apple 9 1 2

橙色4

lemon 12

然而，当我运行时：

IEnumerable<Product> union = store1.Union(store2);

我明白了：

Apple 9 1

橙色4

lemon 12

当我运行时：

IEnumerable<Product> union = store1.Concat(store2);

我明白了：

Apple 9 1

橙色4

Apple 9 2

lemon 12

提前感谢您的帮助。

Answer 1

        //
        // Summary:
        //     Produces the set union of two sequences by using the default equality comparer.
        //
        // Parameters:
        //   first:
        //     An System.Collections.Generic.IEnumerable<T> whose distinct elements form
        //     the first set for the union.
        //
        //   second:
        //     An System.Collections.Generic.IEnumerable<T> whose distinct elements form
        //     the second set for the union.
        //
        // Type parameters:
        //   TSource:
        //     The type of the elements of the input sequences.
        //
        // Returns:
        //     An System.Collections.Generic.IEnumerable<T> that contains the elements from
        //     both input sequences, excluding duplicates.
        //
        // Exceptions:
        //   System.ArgumentNullException:
        //     first or second is null.
public static IEnumerable<TSource> Union<TSource>(this IEnumerable<TSource> first, IEnumerable<TSource> second);

因为函数

不包括重复项。

所以你必须为Product []

编写联合函数

public static class ProductExtension
{
    public static IEnumerable<T> Union<T>(this IEnumerable<T> store1, IEnumerable<T> store2)
    {
        List<T> List = new List<T>();
        foreach (var item in store2)
        {
            if (store1.Any(n=>n.Equals(item)))
            {
                var obj = store1.First(n => n.Equals(item));
                foreach (System.Reflection.PropertyInfo pi in obj.GetType().GetProperties())
                {
                    object v1 = pi.GetValue(obj, null);
                    object v2 = pi.GetValue(item, null);
                    var value = v1;
                    if (v2 != null && (v1 == null || v1.ToString() == string.Empty) && v1 != v2)
                    {
                        value = v2;
                    }
                    pi.SetValue(obj, value, null);
                }
                List.Add(obj);
            }
            else {
                List.Add(item);
            }            
        }

        foreach (var item in store1) { 
            if(!store2.Any(n=>n.Equals(item))){
                List.Add(item);
            }
        }
        return List.AsEnumerable();
    }
}

Answer 2

我不会依赖于反射，而是在Product类中为它编写自己的功能。像AsCombo或CombineWith方法：

public IEnumerable<Product> AsCombo(Product p)
{
    //if not equal, return both
    if (!Equals(p))
    {
        yield return this;
        yield return p;
        yield break;
    }

    //if equal return the one desired by checking all properties
    yield return new Product //always better to return new instance for linq queries
    { 
        Name = Name, 
        Code = Code, 
        Code1 = Code1 ?? p.Code1, //I give preference to 'this'
        Code2 = Code2 ?? p.Code2  //I give preference to 'this'
    };
}

现在所有标准的Linq查询都应该有效。

var combo = store1.Concat(store2)
                  .GroupBy(x => x)
                  .Where(x => x.Count() == 1)
                  .Select(x => x.Key) //get non duplicated products
                  .Concat(store1.Distinct() //concat them with combined form of duplicated products
                                .Join(store2.Distinct(), x => x, x => x, (x, y) => x.AsCombo(y))
                                .SelectMany(x => x))
                  .ToList();

使用下面的查询会更容易，但是那依赖于Union的实现（假设Union保留第一个或外部序列的副本并丢弃后续/内部序列的重复项）。不推荐。

var combo = store1.Join(store2, x => x, x => x, (x, y) => x.AsCombo(y))
                  .SelectMany(x => x)
                  .Union(store1)
                  .Union(store2)
                  .ToList();

如何从另一个可枚举的c＃中插入一个枚举

2 个答案: