WCF返回JSON及EXT调用WCF

差不多整了一天,这个东东刚开始看,非常好的框架,只是配置麻烦些,声明性编程,一时转不过弯弯来,声明的特性类慢慢一个一个了解吧

目标是EXT调用(POST/GET)WCF,数据均以JSON格式传输,理解框架是代码自动生成三层框架+WCF+EXT,

WCF返回JSON的方式,网上流传些两种方式,经测试都可以达到效果,分别是

第一种,dudu大侠的做法是用特性类声明,如

        [OperationContract, WebInvoke( Method = "POST", ResponseFormat = WebMessageFormat.Json )]
        public IList<BA_MenuEntity> DoWork( BA_MenuEntity e ) {
            return BA_MenuBLL.Singleton.Select( new BA_MenuEntity() { Parent = 0 } );
        }

其中ResponseFormat = WebMessageFormat.Json 这一行就可以,比较简单

第二种,后者成峰大侠的做法是利用配置实现返回JSON,具体操作如下

1,添加配置节点,自定义绑定

      <customBinding>
        <binding name="JsonMapper">
          此处配置相当重要，使用了我们编写的JsonContentTypeMapper类，约定返回值类型是Json
          <webMessageEncoding webContentTypeMapperType="Microsoft.Ajax.Samples.JsonContentTypeMapper, JsonContentTypeMapper, Version=1.0.0.0, Culture=neutral, PublicKeyToken=null">
          </webMessageEncoding>
          <httpTransport manualAddressing="true"/>
        </binding>
      </customBinding>

2,写代码继承自WebContentTypeMapper,并在配置中指定此类的子类

    public class JsonContentTypeMapper : WebContentTypeMapper {
        public override WebContentFormat GetMessageFormatForContentType( string contentType ) {
            return WebContentFormat.Json;
        }
    }

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个人比较喜欢前者,dudu大侠的做法,简单些

遇到的其它的问题

1,EXT调用WCF时的传参,如上边的代码,DoWork接收一个自定义实体,该自定义实体比较娇情,它继承了其它的父类,so,一直无法正确返回JSON数据,最后得知:父类同样需要[DataContract]标注,如果该父类有虚属性,则子类必须指定override或new关键字,虽然VS可以编译通过,但WCF无法返回数据

2,出现method not allowed,这个简单,多半是POST/GET方法不对应,如客户端GET请求,而服务器端指定了仅支持POST就会报这个错

3,EXT的AJAX调用WCF时,比jQuery简单些,比dudu那篇简单,起码没有总结部分的3,5(1,2,3是必须的),注意使用EXT的jsonData配置,代码如下

    var webpost = { "__type": "BA_MenuEntity:#Tstring.Core.Entity", "CnName": "信息发布", "EnName": "Information", "ExGUID": null, "ExInclude": false, "ExReamrk": null, "Ext": null, "ID": 1, "Name": null, "ExGuid": "9345E846-42B7-4811-9C0A-DE45257930FE", "Parent": 0, "Remark": "", "Url": "" };
    Ext.Ajax.request({
        url: '/service/defaultservice.svc/DoWork',
        //params: {},
        jsonData: webpost,
        method: 'POST',
        success: function(response, options) {
            var responseArray = Ext.util.JSON.decode(response.responseText);
            if (responseArray != null) {
                Ext.Msg.alert('信息', response.responseText, true);
            }
            else {
                Ext.Msg.alert('失败', '调用失败');
            }
        },
        failure: function(conn, response, options) {
            Ext.Msg.alert('失败', conn.statusText);
        }
    });

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也总结下,(代码环境:visual studio 2008,.NET3.5,EXT3.0)

1,新建ASP.NET WEB应用程序,添加EXT目录

2,新建"启用了AJAX的WCF服务",添加相关的特性类,代码看下来类似如下

namespace wcf2.Service {
    [ServiceContract( Namespace = "" )]
    [AspNetCompatibilityRequirements( RequirementsMode = AspNetCompatibilityRequirementsMode.Allowed )]
    public class DefaultService {

        [OperationContract, WebInvoke( Method = "POST", ResponseFormat = WebMessageFormat.Json )]
        public IList<BA_MenuEntity> DoWork( BA_MenuEntity e ) {
            return BA_MenuBLL.Singleton.Select( new BA_MenuEntity() { Parent = 0 } );
        }

        [OperationContract, WebInvoke( Method = "POST", ResponseFormat = WebMessageFormat.Json )]
        public string GetMenu() {
            return "HELLO WORLD!";
        }
    }
}

3,EXT调用WCF进行测试

最后也有个疑问:

WCF返回JSON,怎么再具体操作它返回的一些数据?如下代码所示:那个"d"是干嘛滴?怎么配置成别的名称?服务器端返回的类型为IList<BA_MenuEntity>,还有那个"__type":"BA_MenuEntity:#Tstring.Core.Entity",

{"d":[{"__type":"BA_MenuEntity:#Tstring.Core.Entity","CnName":"信息发布","EnName":"Information","ExGUID":null,"ExInclude":false,"ExReamrk":null,"Ext":null,"ID":1,"Name":null,"ExGuid":"9345E846-42B7-4811-9C0A-DE45257930FE","Parent":0,"Remark":"","Url":""}]}

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2009-10-22后记

知道那个“d”是干嘛滴了,不过一直没有更新,为了以后复习,备注在此

        [OperationContract( Name = "DoWork4" )]
        [WebInvoke(
            Method = "POST",
            BodyStyle = WebMessageBodyStyle.Wrapped,
            RequestFormat = WebMessageFormat.Json,
            ResponseFormat = WebMessageFormat.Json,
            UriTemplate = "/DoWork4" )]
        [return: MessageParameter( Name = "e" )]
        public Person DoWork4( [MessageParameter( Name = "e" )] Person e ) {
            // 在此处添加操作实现
            return new Person() { Age = 100, Name = "远始天尊" };
        }

加重的两行,[return: MessageParameter( Name = "e" )]这一行影响返回JSON对象时的名称,就是传说中的"d",此时被声明为"e"了

后一行        public Person DoWork4( [MessageParameter( Name = "e" )] Person e ) {

这个影响的是方法接收JSON对象的参数名称,也被改为"e"了,呵呵