我写了一个关于浏览器启动的网络服务工作正常。我在这个webservice中传递一个客户端ID,然后返回一个包含客户端名称的字符串,我们传递的字符串如下:http://prntscr.com/8c1g9z
我创建服务的代码是:
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.ServiceModel;
using System.ServiceModel.Activation;
using System.ServiceModel.Web;
namespace RESTService.Lib
{
[ServiceContract(Name = "RESTDemoServices")]
public interface IRESTDemoServices
{
[OperationContract]
[WebGet(UriTemplate = "/Client/{id}", BodyStyle = WebMessageBodyStyle.Bare)]
string GetClientNameById(string Id);
}
[ServiceBehavior(InstanceContextMode = InstanceContextMode.Single, ConcurrencyMode = ConcurrencyMode.Single, IncludeExceptionDetailInFaults = true)]
[AspNetCompatibilityRequirements(RequirementsMode = AspNetCompatibilityRequirementsMode.Allowed)]
public class RestDemoServices:IRESTDemoServices
{
public string GetClientNameById(string Id)
{
return ("Le nom de client est Jack et id est : " +Id);
}
}
}
但我无法消耗它。我的代码是:
using System;
using System.Collections.Generic;
using System.Linq;
using System.Web;
using System.Web.UI;
using System.Web.UI.WebControls;
using System.Net.Http;
using System.Net;
using System.IO;
namespace ConsumerClient
{
public partial class _Default : System.Web.UI.Page
{
protected void Page_Load(object sender, EventArgs e)
{
System.Net.HttpWebRequest webrequest = (HttpWebRequest)System.Net.WebRequest.Create("http://localhost:8000/DEMOService/Client/156");
webrequest.Method = "POST";
webrequest.ContentType = "application/json";
webrequest.ContentLength = 0;
Stream stream = webrequest.GetRequestStream();
stream.Close();
string result;
using (WebResponse response = webrequest.GetResponse()) //It gives exception at this line liek this http://prntscr.com/8c1gye
{
using (StreamReader reader = new StreamReader(response.GetResponseStream()))
{
result = reader.ReadToEnd();
Label1.Text = Convert.ToString(result);
}
}
}
}
}
我得到一个像这样的例外http://prntscr.com/8c1gye如何使用Web服务。有人可以帮助我吗?
答案 0 :(得分:13)
异常非常明确 - 如果您想从REST服务检索数据,则无法使用POST,除非它允许。您应该使用GET
代替POST
,或者只是不要更改request.Method
。默认情况下,它是GET
。
你不需要做任何特别的事情来消费" REST服务 - 基本上它们就像任何其他URL一样工作。 HTTP POST
谓词表示您希望创建新资源或发布表单数据。要检索资源(页面,API响应等),请使用GET。
这意味着您可以使用任何与HTTP相关的.NET类来调用REST服务 - HttpClient(首选),WebClient或原始HttpWebRequest。
SOAP服务使用POST来获取和发送数据,现在每个人(包括SOAP的创建者)都认为这是设计错误。
修改强>
为清楚起见,使用GET
表示没有内容,也不需要或不允许与内容相关的标题或操作。它与下载任何HTML页面相同:
var url="http://localhost:8000/DEMOService/Client/156";
var webrequest = (HttpWebRequest)System.Net.WebRequest.Create(url);
using (var response = webrequest.GetResponse())
using (var reader = new StreamReader(response.GetResponseStream()))
{
var result = reader.ReadToEnd();
Label1.Text = Convert.ToString(result);
}
您甚至可以将URL直接粘贴到浏览器以获得相同的行为
答案 1 :(得分:0)
这个api是如何在c#
中使用Rest Web Services的完美示例// http://localhost:{portno} / api / v1 / youractionname?UserName = yourusername& Passowrd = yourpassword [HttpGet]
[ActionName("Youractionname")]
public Object Login(string emailid, string Passowrd)
{
if (emailid == null || Passowrd == null)
{
geterror gt1 = new geterror();
gt1.status = "0";
gt1.msg = "All field is required";
return gt1;
}
else
{
string StrConn = ConfigurationManager.ConnectionStrings["cn1"].ConnectionString;
string loginid = emailid;
string Passwrd = Passowrd;
DataTable dtnews = new DataTable();
SqlConnection con = new SqlConnection(StrConn);
con.Open();
SqlCommand cmd = new SqlCommand("sp_loginapi_app", con);
cmd.CommandType = CommandType.StoredProcedure;
SqlParameter p1 = new SqlParameter("@emailid", loginid);
SqlParameter p2 = new SqlParameter("@password", Passowrd);
SqlDataAdapter da = new SqlDataAdapter(cmd);
cmd.Parameters.Add(p1);
cmd.Parameters.Add(p2);
da.Fill(dtnews);
if (dtnews.Rows[0]["id"].ToString() == "-1")
{
geterror gt1 = new geterror();
gt1.status = "0";
gt1.msg = "Invalid Username or Password";
con.Close();
return gt1;
}
else
{
dtmystring.Clear();
dtmystring.Columns.Add(new DataColumn("id", typeof(int)));
dtmystring.Columns.Add(new DataColumn("Name", typeof(string)));
dtmystring.Columns.Add(new DataColumn("Password", typeof(string)));
dtmystring.Columns.Add(new DataColumn("MobileNo", typeof(string)));
dtmystring.Columns.Add(new DataColumn("Emailid", typeof(string)));
DataRow drnew = dtmystring.NewRow();
drnew["id"] = dtnews.Rows[0]["id"].ToString();
drnew["Name"] = dtnews.Rows[0]["Name"].ToString();
drnew["Password"] = dtnews.Rows[0]["Password"].ToString();
drnew["MobileNo"] = dtnews.Rows[0]["MobileNo"].ToString();
drnew["Emailid"] = dtnews.Rows[0]["emailid"].ToString();
dtmystring.Rows.Add(drnew);
gt.status = "1";
gt.msg = dtmystring;
con.Close();
return gt;
}
}
}