如何将 Object 作为 c# controller 中的字符串列表传递？

Question

public class abcController : ApiController
{
    public IEnumerable<string> Get()
    {
        return new string[] { "A","B","C"};
    }

}

我想在字符串数组中传递包含名称、id 和图像的 object。如何做到这一点？ 请帮助我.. 还告诉我如何访问此 object 中的任何特定元素..

Answer 1

我想这就是你要找的。 我已经把它放在控制台应用程序中。 您可以将逻辑复制到您的 controller

        public class ImageInformation
        {
            public ImageInformation(string name, string id, string image)
            {
                Name = name;
                Id = id;
                Image = image;
            }
            public string Name { get; private set; }
            public string Id { get; private set; }
            public string Image { get; private set; }
        }

        public static IEnumerable<ImageInformation> Get()
        {
            return new List<ImageInformation> { 
               new ImageInformation("A", "1", "A.jpg"),
               new ImageInformation("B", "2", "B.jpg"),
               new ImageInformation("C", "3", "C.jpg"),
            };
        }

        public static void Main()
        {
            var images = Get();
            var imageA = images.First(x => x.Name == "A");
            var imageAName = imageA.Name;
            var imageAId = imageA.Id;
            var imageAImage = imageA.Image;

            Console.WriteLine(imageAName);
            Console.WriteLine(imageAId);
            Console.WriteLine(imageAImage);
        }

Answer 2

您可以做的是使用 XML 序列化。

What XML Serializing does is it converts an object to XML format (which is a string) and then when you want to get the object back you can use XML Deserializer to Deserialize the XML string format to the original object.

例如，您的 object 可以是包含字符串名称、int ID、字符串 imagePath 的 class，我们称之为 class 人。

Class

[System.Serializable]
public class person
{
   public string name { get; set; }
   public int id { get; set; }
   public string imagePath { get; set; }
}

Class 备注

• 确保 class 具有 system.Serializable 属性。
• 确保您拥有的所有字段和属性都是公开的（否则程序将崩溃。）。

XML 管理器

• 确保您具有以下命名空间，否则您将无法使用列出的任何方法

using System.Xml.Serialization;

XML_Manager class 我们将把我们的方法放在：-

public static class XML_Manager
    {
        // The Serializing Method
        public static string serializeObject<T>(T toSerialize)
        {
            // Create the result string variable
            string result = "";

            // Create a new StringWriter that will carry the serialization and dispose it as soon as it's done it's job
            using(StringWriter sw = new StringWriter())
            {
                // Create the serializer that will serialize the object of type (T)
                XmlSerializer xs = new XmlSerializer(typeof(T));

                // Serialize the object and put the result on the StringWriter (sw)
                xs.Serialize(sw, toSerialize);

                // Exctract the result from the StringWriter (sw) by converting it to string
                result = sw.ToString();
            }

            // Return the result
            return result;
        }

        // The Deserializing Method
        public static T deserializeObject<T>(string XmlString)
        {
            // Create the result variable of type (T)
            T Result;

            // Create a new StringReader that will carry the XmlString and dispose it as soon as it's done it's job
            using (StringReader sr = new StringReader(XmlString))
            {
                // Create the serializer that will deserialize the object of type (T)
                XmlSerializer xs = new XmlSerializer(typeof(T));

                // Deserialize the XmlString and extract the object
                Result = (T)xs.Deserialize(sr);
            }

            // Return the result
            return Result;
        }
    }

使脚本工作

            // Create a new person
            person p = new person();

            // Set the name
            p.name = "John";

            // Set the ID
            p.id = 2345;

            // Set the Image Path
            p.imagePath = "desktop";

            // the xml serialized string
            string xmlString = XML_Manager.serializeObject(p);

            // the deserialized person
            person deserializePerson = XML_Manager.deserializeObject<person>(xmlString);

            // Show the serialized string
            MessageBox.Show(xmlString);

            // Show the deserialized object
            MessageBox.Show($"Name: {deserializePerson.name}, Id: {deserializePerson.id}, Image Path: {deserializePerson.imagePath}");