我正在尝试创建JSON WCF Web服务。 我真的完全不了解整个过程！我正在连接到服务器上的MySQL数据库。 所以我有以下代码： 我的界面 -
[ServiceContract]
public interface IService1
{
[OperationContract]
[WebInvoke(Method = "GET", ResponseFormat = WebMessageFormat.Json, RequestFormat = WebMessageFormat.Json, BodyStyle = WebMessageBodyStyle.Wrapped, UriTemplate = "/GetAllResources")]
List<Resources> GetAllResources();
[OperationContract]
[WebInvoke(Method = "POST", ResponseFormat = WebMessageFormat.Json, RequestFormat = WebMessageFormat.Json, BodyStyle = WebMessageBodyStyle.Wrapped, UriTemplate = "/AddRoom")]
void AddRoom(string location, string name);
...
我的服务 -
[ScriptService]
public class Service1 : IService1
{
[WebMethod]
[ScriptMethod(ResponseFormat = ResponseFormat.Json)]
public void AddRoom(string location, string name)
{
String conString = System.Configuration.ConfigurationManager.ConnectionStrings["MyDatabaseConnectionString"].ConnectionString;
using (MySqlConnection cnn = new MySqlConnection(conString))
{
cnn.Open();
String sql = String.Format("INSERT INTO rooms(roomLocation, roomName) VALUES ({0}, {1});", location, name);
MySqlCommand cmd = new MySqlCommand(sql, cnn);
//doesn't return any rows
cmd.ExecuteNonQuery();
}
}
[WebMethod]
[ScriptMethod(ResponseFormat = ResponseFormat.Json)]
public List<Resources> GetAllResources()
{
String conString = System.Configuration.ConfigurationManager.ConnectionStrings["MyDatabaseConnectionString"].ConnectionString;
List<Resources> al = new List<Resources>();
using (MySqlConnection cnn = new MySqlConnection(conString))
{
cnn.Open();
String sql = String.Format("select * from resources");
MySqlCommand cmd = new MySqlCommand(sql, cnn);
MySqlDataReader reader = cmd.ExecuteReader();
while (reader.Read())
{
al.Add((Resources)reader[0]);
}
return al;
}
}
...
Web配置 -
...
<system.web.extensions>
<scripting>
<webServices>
<jsonSerialization maxJsonLength="5000"/>
</webServices>
</scripting>
</system.web.extensions>
<system.web>
<compilation debug="true" targetFramework="4.5" />
<httpRuntime targetFramework="4.5"/>
</system.web>
<system.serviceModel>
<services>
<service name="RoomBookingService.Service1" behaviorConfiguration="RoomBookingServiceBehavior">
<endpoint address="../Service1.svc"
binding="webHttpBinding"
contract="RoomBookingService.IService1"
behaviorConfiguration="webBehaviour" />
</service>
</services>
<behaviors>
<serviceBehaviors>
<behavior name="RoomBookingServiceBehavior">
<!-- To avoid disclosing metadata information, set the values below to false before deployment -->
<serviceMetadata httpGetEnabled="true" httpsGetEnabled="true"/>
<!-- To receive exception details in faults for debugging purposes, set the value below to true. Set to false before deployment to avoid disclosing exception information -->
<serviceDebug includeExceptionDetailInFaults="false"/>
</behavior>
</serviceBehaviors>
<endpointBehaviors>
<behavior name="webBehaviour">
<webHttp automaticFormatSelectionEnabled="true"/>
</behavior>
</endpointBehaviors>
</behaviors>
...
这是对的吗？我可以使用哪些工具来测试服务？我已将它放到服务器上并尝试下载一些测试工具，但是他们没有给我任何错误，只是因为它没有返回JSON？！
我将创建一个与该服务交谈的Android应用程序，但由于这也是一个学习曲线，我想知道我的服务在添加另一层复杂性之前正常工作。
对我的代码或我的问题的任何帮助或评论将不胜感激。 谢谢你的时间
答案 0 :(得分：0)
我设法让它工作： 这是我的代码： 的合同：
namespace RoomBookingService
{
// NOTE: You can use the "Rename" command on the "Refactor" menu to change the interface name "IService1" in both code and config file together.
[ServiceContract]
public interface IService1
{
[OperationContract]
[WebGet(ResponseFormat = WebMessageFormat.Json, RequestFormat = WebMessageFormat.Json, BodyStyle = WebMessageBodyStyle.Wrapped, UriTemplate = "GetAllResources")]
String GetAllResources();
<强>服务
[WebMethod]
[ScriptMethod(ResponseFormat = ResponseFormat.Json)]
public String GetAllResources()
{
String conString = System.Configuration.ConfigurationManager.ConnectionStrings["MyDatabaseConnectionString"].ConnectionString;
List<Dictionary<string, object>> tableRows = new List<Dictionary<string, object>>();
Dictionary<string, object> row= new Dictionary<string,object>();
DataTable dt = new DataTable();
System.Web.Script.Serialization.JavaScriptSerializer serializer =
new System.Web.Script.Serialization.JavaScriptSerializer();
try
{
using (MySqlConnection cnn = new MySqlConnection(conString))
{
cnn.Open();
String sql = String.Format("select resourceID, resourceName, resourceDesc, roomID from resources");
MySqlCommand cmd = new MySqlCommand(sql, cnn);
MySqlDataReader reader = cmd.ExecuteReader();
dt.Load(reader);
foreach (DataRow dr in dt.Rows)
{
row = new Dictionary<String, Object>();
foreach (DataColumn col in dt.Columns)
{
row.Add(col.ColumnName, dr[col]);
}
tableRows.Add(row);
}
return serializer.Serialize(tableRows);
}
}
catch (Exception ex)
{
return ex.ToString();
}
}
<强> WebConfig
<system.web.extensions>
<scripting>
<webServices>
<jsonSerialization maxJsonLength="5000"/>
</webServices>
</scripting>
</system.web.extensions>
<system.web>
<compilation debug="true" targetFramework="4.5" />
<httpRuntime targetFramework="4.5"/>
</system.web>
<system.serviceModel>
<services>
<service name="RoomBookingService.Service1" behaviorConfiguration="RoomBookingServiceBehavior">
<endpoint address=""
binding="webHttpBinding"
contract="RoomBookingService.IService1"
behaviorConfiguration="webBehaviour" />
</service>
</services>
<behaviors>
<serviceBehaviors>
<behavior name="RoomBookingServiceBehavior">
<!-- To avoid disclosing metadata information, set the values below to false before deployment -->
<serviceMetadata httpGetEnabled="true" httpsGetEnabled="true"/>
<!-- To receive exception details in faults for debugging purposes, set the value below to true. Set to false before deployment to avoid disclosing exception information -->
<serviceDebug includeExceptionDetailInFaults="false"/>
</behavior>
</serviceBehaviors>
<endpointBehaviors>
<behavior name="webBehaviour">
<webHttp automaticFormatSelectionEnabled="true"/>
</behavior>
</endpointBehaviors>
除了它的工作之外，还不是绝对清楚！所以我会说： - ）
谢谢！