Question

我有这个问题

使用获取

我想发送一篇http帖子

这是我的Android类和请求方法

    public static String readUrl(String url, ArrayList<NameValuePair> params) {
    try {
        HttpClient client = new DefaultHttpClient();
        HttpPost method = new HttpPost(url);

        if (params != null) {
            method.setEntity(new UrlEncodedFormEntity(params));

        }

        HttpResponse response = client.execute(method);

        InputStream inputStream = response.getEntity().getContent();
        String result = convertInputStreamToString(inputStream);

        return result;
    }
    catch (ClientProtocolException e) {
        e.printStackTrace();
    }
    catch (IOException e) {
        e.printStackTrace();
    }

    return null;
}

这是请求代码

                                ArrayList<NameValuePair> params = new ArrayList<NameValuePair>();
                                params.add(new BasicNameValuePair("region","1"));
                                params.add(new BasicNameValuePair("area","1"));
                                params.add(new BasicNameValuePair("sector","1"));
                                params.add(new BasicNameValuePair("address",edtReqAddres.getText().toString()));
                                params.add(new BasicNameValuePair("mobile",edtMobileNumber.getText().toString()));

                                params.add(new BasicNameValuePair("username","test"));
                                params.add(new BasicNameValuePair("message", edtSubject.getText().toString()));
                                params.add(new BasicNameValuePair("compid","0"));
                                params.add(new BasicNameValuePair("geo","1"));


                                text = webservice.readUrl("http://192.168.1.102:81/api/products", params);

而且这也是我的结果：（

{＆＃34;消息＆＃34;：＆＃34;未找到与请求URI匹配的HTTP资源＆＃39; http://192.168.1.102:81/api/products＆＃39;。＆＃34;}

这是我的WebApi（DotNet）

    [System.Web.Http.AcceptVerbs("GET", "POST")]
    public string GetProductById(int region, int area, int sector, string address, string mobile, string username, string message, int compid, string geo)
    {
      fddService.mobService service=new mobService();
      return service.NewMessage(region, area, sector, address, mobile, "", message, compid, "");
    }

Answer 1

如果在Visual Studio中使用ASP.NET WebAPI，则可以添加新的Web API Controller Class（v2.1），然后您将拥有以下默认代码：

public class ProductsController : ApiController
    {
        // GET api/<controller>
        public IEnumerable<string> Get()
        {
            return new string[] { "value1", "value2" };
        }

    // GET api/<controller>/5
    public string Get(int id)
    {
        return "value";
    }

    // POST api/<controller>
    public void Post([FromBody]string value)
    {
    }

    // PUT api/<controller>/5
    public void Put(int id, [FromBody]string value)
    {
    }

    // DELETE api/<controller>/5
    public void Delete(int id)
    {
    }
}

然后，假设您有一个DTO类，例如产品，您可以按以下方式自定义Web API：

        // POST: api/Products
        [ResponseType(typeof(Product))]
        public async Task<IHttpActionResult> PostProduct(Product product)
        {
            if (!ModelState.IsValid)
            {
                return BadRequest(ModelState);
            }

            db.Products.Add(product);
            await db.SaveChangesAsync();

            return CreatedAtRoute("DefaultApi", new { id = product.Id }, product);
        }

您可以在Learn About ASP.NET Web API

找到更多信息

Answer 2

所以最后我得到了答案在客户端您应该将此行添加到您的代码中以设置defult内容类型

httpPost.addHeader("Content-type", "application/x-www-form-urlencoded");

在后端服务器（WebApi）上应创建 ModelView 以发送参数，并在输入参数的开头添加[FromBody]，如下所示< / p>

public string postreq([FromBody] RequestViewModel model)
{
    fddService.mobService service = new mobService();
    if (model.geo == "1")
        return service.NewMessage(Convert.ToInt32(model.region), Convert.ToInt32(model.area),
            Convert.ToInt32(model.sector), model.address, model.mobile, "", model.message, Convert.ToInt32(model.compid), "");
    else
        return service.NewMessage(Convert.ToInt32(model.region), Convert.ToInt32(model.area),
            Convert.ToInt32(model.sector), model.address, model.mobile, "", model.message, Convert.ToInt32(model.compid), model.geo);
}
public class RequestViewModel
{
    public string region { get; set; }
    public string area { get; set; }
    public string sector { get; set; }
    public string address { get; set; }
    public string mobile { get; set; }
    public string username { get; set; }
    public string message { get; set; }
    public string compid { get; set; }
    public string geo { get; set; }
}

如果你有{"Message":"An error has occurred."}：

确保您的Android客户端正确地使用正确的POST代码传递参数。

谢谢：Earthling POST data from Android to Web API returns 404

发送http帖子到dot net webapi

2 个答案: