为什么我的WCF Rest服务方法的参数总是为空?....我确实访问了服务的方法,我确实得到了wcf方法返回的字符串,但参数仍为null。
运营合同:
[OperationContract]
[WebInvoke(UriTemplate = "AddNewLocation",
Method="POST",
BodyStyle = WebMessageBodyStyle.WrappedRequest,
ResponseFormat = WebMessageFormat.Json,
RequestFormat = WebMessageFormat.Json)]
string AddNewLocation(NearByAttractions newLocation);
AddNewLocation方法的实现
public string AddNewLocation(NearByAttractions newLocation)
{
if (newLocation == null)
{
//I'm always getting this text in my logfile
Log.Write("In add new location:- Is Null");
}
else
{
Log.Write("In add new location:- " );
}
//String is returned even though parameter is null
return "59";
}
客户代码:
WebClient clientNewLocation = new WebClient();
clientNewLocation.Headers[HttpRequestHeader.ContentType] = "application/json";
JavaScriptSerializer js = new JavaScriptSerializer();
js.MaxJsonLength = Int32.MaxValue;
//Serialising location object to JSON
string serialLocation = js.Serialize(newLocation);
//uploading JSOn string and retrieve location's ID
string jsonLocationID = clientNewLocation.UploadString(GetURL() + "AddNewLocation", serialLocation);
我也在我的客户端尝试过这段代码,但仍然得到一个空参数
DataContractJsonSerializer ser = new DataContractJsonSerializer(typeof(NearByAttractions));
MemoryStream ms = new MemoryStream();
ser.WriteObject(ms, newLocation);
String json = Encoding.UTF8.GetString(ms.ToArray());
WebClient clientNewLocation = new WebClient();
clientNewLocation.Headers[HttpRequestHeader.ContentType] = "application/json";
string r = clientNewLocation.UploadString(GetURL() + "AddNewLocation", json);
Console.Write(r);
然后我也将 BodyStyle 选项更改为“ Bare ”,但后来出现以下错误(包含两个客户端代码):
远程服务器返回错误:(400)错误请求。
请帮忙吗?感谢
编辑1: 我的GetUrl()方法从Web配置文件加载Web服务IP地址,并返回Uri类型的对象
private static Uri GetURL()
{
Configuration config = WebConfigurationManager.OpenWebConfiguration("~/web.config");
string sURL = config.AppSettings.Settings["serviceURL"].Value;
Uri url = null;
try
{
url = new Uri(sURL);
}
catch (UriFormatException ufe)
{
Log.Write(ufe.Message);
}
catch (ArgumentNullException ane)
{
Log.Write(ane.Message);
}
catch (Exception ex)
{
Log.Write(ex.Message);
}
return url;
}
存储在Web配置中的服务地址如下:
<appSettings>
<add key="serviceURL" value="http://192.168.2.123:55666/TTWebService.svc/"/>
</appSettings>
这是我的NearByAttraction类定义的方式
[DataContractAttribute]
public class NearByAttractions
{
[DataMemberAttribute(Name = "ID")]
private int _ID;
public int ID
{
get { return _ID; }
set { _ID = value; }
}
[DataMemberAttribute(Name = "Latitude")]
private string _Latitude;
public string Latitude
{
get { return _Latitude; }
set { _Latitude = value; }
}
[DataMemberAttribute(Name = "Longitude")]
private string _Longitude;
public string Longitude
{
get { return _Longitude; }
set { _Longitude = value; }
}
答案 0 :(得分:4)
你似乎走在正确的轨道上。您需要Bare
正文样式,否则您需要将输入的序列化版本包装在另一个JSON对象中。第二个代码应该可行 - 但是如果没有关于如何设置服务以及GetURL()
返回的更多信息,我们只能猜测。
找出要发送到WCF REST服务的内容的一种方法是使用WCF客户端本身 - 使用WebChannelFactory<T>
类,然后使用Fiddler等工具查看它发送的内容。以下示例为SSCCE,可显示您的方案正常运行。
public class StackOverflow_15786448
{
[ServiceContract]
public interface ITest
{
[OperationContract]
[WebInvoke(UriTemplate = "AddNewLocation",
Method = "POST",
BodyStyle = WebMessageBodyStyle.Bare,
ResponseFormat = WebMessageFormat.Json,
RequestFormat = WebMessageFormat.Json)]
string AddNewLocation(NearByAttractions newLocation);
}
public class NearByAttractions
{
public double Lat { get; set; }
public double Lng { get; set; }
public string Name { get; set; }
}
public class Service : ITest
{
public string AddNewLocation(NearByAttractions newLocation)
{
if (newLocation == null)
{
//I'm always getting this text in my logfile
Console.WriteLine("In add new location:- Is Null");
}
else
{
Console.WriteLine("In add new location:- ");
}
//String is returned even though parameter is null
return "59";
}
}
public static void Test()
{
string baseAddress = "http://" + Environment.MachineName + ":8000/Service";
WebServiceHost host = new WebServiceHost(typeof(Service), new Uri(baseAddress));
host.Open();
Console.WriteLine("Host opened");
Console.WriteLine("Using WCF-based client (WebChannelFactory)");
var factory = new WebChannelFactory<ITest>(new Uri(baseAddress));
var proxy = factory.CreateChannel();
var newLocation = new NearByAttractions { Lat = 12, Lng = -34, Name = "56" };
Console.WriteLine(proxy.AddNewLocation(newLocation));
Console.WriteLine();
Console.WriteLine("Now with WebClient");
DataContractJsonSerializer ser = new DataContractJsonSerializer(typeof(NearByAttractions));
MemoryStream ms = new MemoryStream();
ser.WriteObject(ms, newLocation);
String json = Encoding.UTF8.GetString(ms.ToArray());
WebClient clientNewLocation = new WebClient();
clientNewLocation.Headers[HttpRequestHeader.ContentType] = "application/json";
string r = clientNewLocation.UploadString(baseAddress + "/AddNewLocation", json);
Console.WriteLine(r);
}
}
答案 1 :(得分:3)
解决了,谢谢
我将BodyStyle更改为“Bare”所以我的服务界面如下:
[OperationContract]
[WebInvoke(UriTemplate = "AddNewLocation",
Method="POST",
BodyStyle = WebMessageBodyStyle.Bare,
ResponseFormat = WebMessageFormat.Json,
RequestFormat = WebMessageFormat.Json)]
string AddNewLocation(NearByAttractions newLocation);
然后按如下方式实现我的客户端:
MemoryStream ms = new MemoryStream();
DataContractJsonSerializer serialToUpload = new DataContractJsonSerializer(typeof(NearByAttractions));
serialToUpload.WriteObject(ms, newLocation);
WebClient client = new WebClient();
client.Headers.Add(HttpRequestHeader.ContentType, "application/json");
client.UploadData(GetURL() + "AddNewLocation", "POST", ms.ToArray());
我使用了WebClient.UploadData而不是UploadString。