“远程服务器返回错误：（404）”在桌面应用程序框架 3.5 [已解决]

Question

我正在为我的桌面应用程序使用 API。该应用程序运行在 dot.net 框架 3.5.

我有一个可以完美运行的 API，但是当我从桌面应用程序调用 API 时，出现错误：

“远程服务器返回错误：（404）”

但是我的API运行没有问题。我有两个参数，将在 URL 中传递。

这是我的代码：

string url = "http://localhost:58167/api/Project/";
string data = "65354/19216882";
string response;
WebClient client = new WebClient();
{
     client.Encoding = Encoding.UTF8;
     response = client.downloadstring(url, "POST", data);
}

我该如何解决这个问题？

编辑： 这是说明 API 外观的代码：

    // GET api/Project
    public string GetProjects(string key, string IP)
    {
        string sql = "";
        string en = "";
        if (IP == "19216")
            sql = "garbage1";
        if (IP == "19882")
            sql = " garbage2";
        if (IP == "181249")
            sql = " garbage3";
        if (IP == "85206")
            sql = " garbage4";
        if (IP == "87249")
            sql = " garbage5";

        en = CryptorEngine.Encrypt(sql, key);
        return en;
    }

更新解决的代码：

string url = "http://localhost:58167/api/Project/";
string data = "65354/19216882";
string response;
WebClient client = new WebClient();
{
     client.Encoding = Encoding.UTF8;
     response = client.downloadstring(url, "POST", data);
}

Answer 1

尝试将属性路由添加到您的 API

 Route[("~/api/Project/{key}/{ip}")]
 public string GetProjects(string key, string IP)

你的代码

string url = "http://localhost:58167/api/Project/65354/19216882";
string response;
WebClient client = new WebClient();
{
     client.Encoding = Encoding.UTF8;
     response = client.UploadString(url, "GET");
}

如果您无法控制 api

你可以使用现有的动作

Route[("~/api/Project")]
public string GetProjects(string key, string IP)

代码

string url = "http://localhost:58167/api/Project?key=65354&IP=19216882";
    
string response;
WebClient client = new WebClient();
{
     client.Encoding = Encoding.UTF8;
     response = client.UploadString(url, "GET");
}

如果您仍然需要使用 POST，则必须修复 acion

public class ViewModel
{
 public  string Key {get; set;}
  public string IP  {get; set;}
}

 Route[("~/api/Project")]
 public string GetProjects([FromBody]ViewModel model)

Answer 2

谢谢大家。我得到了解决，并更新了有问题的更新代码。只需将 uploadstring 更改为 downloadstring 。和宾果……它有效。 –

更新解决的代码：

string url = "http://localhost:58167/api/Project/";
string data = "65354/19216882";
string response;
WebClient client = new WebClient();
{
     client.Encoding = Encoding.UTF8;
     response = client.downloadstring(url, "POST", data);
}