2014-09-11 53 views
0
public FooMessageType Foo(string name, string password) 
{ 
    SqlConnection con = new SqlConnection(WebConfigurationManager.ConnectionStrings["sample"].ToString()); 
    con.Open(); 
    SqlCommand cmd = new SqlCommand("Select * from Employee where username='" + name + "' and password='" + password + "'", con); 
    SqlDataReader rd = cmd.ExecuteReader(); 
    if (rd.Read()) 
    { 
     return new FooMessageType { Name = name, Password = password, msg = "welcome" }; 
    } 
    else 
    { 
     return new FooMessageType { Name = name, Password = password, msg = "error" }; 
    } 
} 

[OperationContract] 
[WebInvoke(Method = "POST", 
    ResponseFormat = WebMessageFormat.Json, 
    RequestFormat = WebMessageFormat.Json, 
    BodyStyle = WebMessageBodyStyle.WrappedRequest, UriTemplate = "aw")] 
FooMessageType Foo(string name, string password); 

[DataContract] 
public class FooMessageType 
{ 
    [DataMember(Name = "name")] 
    public string Name { get; set; } 
    [DataMember(Name = "password")] 
    public string Password { get; set; } 
    [DataMember(Name = "msg")] 
    public string msg { get; set; } 
} 

这是请求数据如何格式化JSON显示和获得响应

{ "name": "123","password": "123"} 

这是响应数据

{"msg":"error","name":"123","password":"123"} 

我想要什么请求

{LoginRequest:{"username": "123","password":"123"}} 

和响应

{LoginResponse:{"responseCode":"1","responseMessage":"" 
userDetails: {"firstname":"","lastname": ""}}} 

任何人都可以帮我创建一个类来显示这些数据,得到的回应?我是创建wcf的新手。

+0

这是无效的JSON。 – SLaks 2014-09-11 03:10:41

+0

我更新了它... – newb 2014-09-11 03:15:11

+0

所以你问如何添加一个属性到一个类?你有什么尝试?你有什么问题? – SLaks 2014-09-11 03:21:44

回答

0
class OuterClass { 
    public InnerClass MyProperty { get; set; } 
} 

class InnerClass { ... }