Linq Distinct 不按照论坛和Microsoft示例运行
本文关键字:Microsoft 运行 论坛 Distinct Linq | 更新日期: 2023-09-27 18:21:16
using System;
using System.Collections.Generic;
using System.Linq;
using NUnit.Framework;
namespace UnitTest.Model
{
[TestFixture]
public class SampleEquatableObjectTest
{
[Test]
public void TwoIdenticalUsersComparedEqualTrue()
{
var user1 = new SampleObject { Id = 1, Name = "Test User" };
var user2 = new SampleObject { Id = 1, Name = "Test User" };
Assert.IsTrue(user1.Equals(user2));
}
[Test]
public void TwoDifferentUsersComparedEqualFalse()
{
var user1 = new SampleObject { Id = 1, Name = "Test User 1" };
var user2 = new SampleObject { Id = 2, Name = "Test User 2" };
Assert.IsFalse(user1.Equals(user2));
}
[Test]
public void CollectionOfUsersReturnsDistinctList()
{
var userList = new List<SampleObject>
{
new SampleObject {Id = 1, Name = "Test User"},
new SampleObject {Id = 1, Name = "Test User 1"},
new SampleObject {Id = 2, Name = "Test User 2"}
};
Assert.AreEqual(userList.Count, 3);
var result = userList.Distinct();
Assert.AreEqual(result.Count(), 2);
var multipleTest = (from r in result group r by new { r.Id } into multGroup where multGroup.Count() > 1 select multGroup.Key).Any();
Assert.IsFalse(multipleTest);
}
public class SampleObject : IEquatable<SampleObject>
{
public int Id { get; set; }
public string Name { get; set; }
public bool Equals(SampleObject other)
{
if (ReferenceEquals(this, other))
return true;
if (ReferenceEquals(other, null) || ReferenceEquals(this, null))
return false;
return Id.Equals(other.Id);
}
}
}
}
此测试用例中的非重复方法不会返回非重复列表。计数断言将失败。我查看了其他类似的问题和Microsoft示例,但它们看起来与我在测试中的代码完全相同。任何输入?
您还需要从 Object 类重写 GetHashCode(( 和 Equals 方法。有关更多信息,请参阅此相应的 FXCOP 违规。
然后,您的测试将按预期工作。
public class SampleObject : IEquatable<SampleObject>
{
public int Id { get; set; }
public string Name { get; set; }
public bool Equals(SampleObject other)
{
if (ReferenceEquals(this, other))
return true;
if (ReferenceEquals(other, null) || ReferenceEquals(this, null))
return false;
return Id.Equals(other.Id);
}
public override int GetHashCode()
{
return Id;
}
public override bool Equals(object obj)
{
return Equals(obj as SampleObject);
}
}
您正在实现一个接口,而无需根据接口的规范定义两个成员。 您的代码需要如下所示:
public bool Equals(SampleObject x, SampleObject y)
{
if (ReferenceEquals(x, y))
return true;
if (ReferenceEquals(x, null) || ReferenceEquals(y, null))
return false;
return x.Id.Equals(y.Id);
}
public int GetHashCode(SampleObject obj)
{
public int GetHashCode(SampleObject obj)
{
if (Object.ReferenceEquals(obj, null)) return 0;
int hashId = obj.Id == null ? 0 : obj.Id.GetHashCode();
int hashName = obj.Name == null ? 0 : obj.Name.GetHashCode();
return hashId ^ hashName; // or what ever you want you hash to be, hashID would work just as well.
}
}