如何：根据用例场景将不同类型的对象作为参数传递给方法

Question

我试图不重复代码，所以我创建了一个参数化的 CreateFile 方法。 它的参数之一 myParams 有时可以是 CustomObject_A 类型，而有时它可以是 CustomObject_B 类型。那么我该如何实现呢？

代码下方：

public class CustomObject_A {
    public DataTable dt { get; set; }
    public string prop1 { get; set; }
}

public class CustomObject_B {
    public List<AnotherCustomObject> lst { get; set; }
    public int prop2 { get; set; }
}

public void DoWork_CreateFile_A(string baseUrl)
{
        string requestUri = "api/MyController/CreateFile_A";
        CustomObject_A myParams = new CustomObject_A();

        this.CreateFile(baseUrl, requestUri, myParams);
}

public void DoWork_CreateFile_B(string baseUrl)
{
        string requestUri = "api/MyController/CreateFile_B";
        CustomObject_B myParams = new CustomObject_B();

        this.CreateFile(baseUrl, requestUri, myParams);
}

public async Task CreateFile(string strBaseUrl, string requestUri, someType myParams)
{        
     using (HttpClient client = new HttpClient())
     {
          client.BaseAddress = new Uri(strBaseUrl);
          client.DefaultRequestHeaders.Clear();
          client.DefaultRequestHeaders.Accept.Add(new System.Net.Http.Headers.MediaTypeWithQualityHeaderValue("application/json"));

          string stringMyParams = await Task.Run(() => JsonConvert.SerializeObject(myParams, Newtonsoft.Json.Formatting.Indented,JsonSerializerSettings { ContractResolver = new CamelCasePropertyNamesContractResolver() }));

          var httpContent = new StringContent(stringMyParams, Encoding.UTF8, "application/json");
          HttpResponseMessage res = await client.PostAsync(requestUri, httpContent);

          if (res.IsSuccessStatusCode)
          {
              // Do things
          }
     }
}

Web API 控制器：

public class MyController : ApiController
{
        public bool CreateFile_A(string data)
        {
              CustomObject_A myData = JsonConvert.DeserializeObject<CustomObject_A>(data);
              // Do some stuff here
        }

        public bool CreateFile_B(string data)
        {
             CustomObject_B myData = JsonConvert.DeserializeObject<CustomObject_B>(data);
             // Do some stuff here
        }
}

Answer 1

你所有类似的类都需要继承一个通用接口，然后你可以使用接口作为方法所采用的类型。任何继承该接口的东西都可以传入。

Answer 2

有多种方法可以实现，取决于你想完成什么..

1. 通用方法

   public async Task CreateFile<T>(string strBaseUrl, string requestUri, T myParams)
    { 
       ....
    }
   

2. 界面

    public void DoSomething<T>(T param) where T: IFileCreator
    {
       param.CreateFile();
    }
   
    public interface IFileCreator
    {
        void CreateFile();  
    }
   
    public abstract class FileCreatorBase : IFileCreator
    {
        public virtual void CreateFile()
        {
           //common code
        }
    }
   
    public class ClassA : FileCreatorBase
    {
        public int Id { get; set; }
   
        public override void CreateFile()
        {
            Id.Dump();
        }
    }
   
    public class ClassB : FileCreatorBase
    {
        public string Name { get; set; }    
   
        public override void CreateFile()
        {
           Name.Dump();
        }
    }
   

3. 带虚拟方法的抽象类

        public abstract class FileCreator
        {
            public virtual void CreateFile()
            {
                //common code
            }
        }
   
        public class ClassA : FileCreator
        {
           public int Id { get; set; }
   
           public override void CreateFile()
           {
              Id.Dump();
           }
        }
   
        public class ClassB : FileCreator
        {
            public string Name { get; set; }
   
            public override void CreateFile()
            {
                Name.Dump();
            }
        }
   

4. 带有抽象方法的抽象类

    public abstract class FileCreator
    {
        public abstract void CreateFile();  
    }
   
    public class ClassA : FileCreator
    {
       public int Id { get; set; }
   
       public override void CreateFile()
       {
          Id.Dump();
       }
    }
   
    public class ClassB : FileCreator
    {
        public string Name { get; set; }
   
        public override void CreateFile()
        {
            Name.Dump();
        }
    }