未找到 Java 类型类 net.sf.json.JSONObject 和 MIME 媒体类型 application/json 的消息正文编写器

A message body writer for Java type, class net.sf.json.JSONObject, and MIME media type, application/json, was not found

提问by sorabh solanki

I am using jersey client for making call to rest webservice.

我正在使用球衣客户端来调用休息网络服务。

My webservice is consuming the json, so i need to put json in making call to my webservice provider.

我的网络服务正在使用 json，所以我需要将 json 放入我的网络服务提供者的调用中。

I am doing it in below way.

我正在按照以下方式进行操作。

    JSONObject object=new JSONObject();
    object.put("name", employee.getName());
    object.put("empId", employee.getEmpId());
    object.put("organizationName", employee.getOrganizationName());


ClientResponse response = service.path("rest").path("vtn").path("addEmplyee")
            .type(MediaType.APPLICATION_JSON_TYPE).post(ClientResponse.class, object);

but i am getting the below exception:

但我收到以下异常：

09:52:01,625 ERROR [[mvc-dispatcher]] Servlet.service() for servlet mvc-dispatcher threw exception com.sun.jersey.api.client.ClientHandlerException: A message body writer for Java type, class net.sf.json.JSONObject, and MIME media type, application/json, was not foundat com.sun.jersey.api.client.RequestWriter.writeRequestEntity(RequestWriter.java:288) at com.sun.jersey.client.urlconnection.URLConnectionClientHandler._invoke(URLConnectionClientHandler.java:204) at com.sun.jersey.client.urlconnection.URLConnectionClientHandler.handle(URLConnectionClientHandler.java:147) at com.sun.jersey.api.client.Client.handle(Client.java:648) at com.sun.jersey.api.client.WebResource.handle(WebResource.java:670) at com.sun.jersey.api.client.WebResource.access$200(WebResource.java:74) at com.sun.jersey.api.client.WebResource$Builder.post(WebResource.java:563) at com.nec.jp.pflow.unc.service.EmployeeService.addEmployee(EmployeeService.java:44) at com.nec.jp.pflow.unc.controller.EmployeeController.addCustomer(EmployeeController.java:29) at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method) at sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:39) at sun.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:25)

09:52:01,625 错误 [[mvc-dispatcher]] servlet mvc-dispatcher 的 Servlet.service() 抛出异常 com.sun.jersey.api.client.ClientHandlerException：Java 类型的消息正文编写器，类 net.sf。未找到 json.JSONObject 和 MIME 媒体类型 application/json在 com.sun.jersey.api.client.RequestWriter.writeRequestEntity(RequestWriter.java:288) 在 com.sun.jersey.client.urlconnection.URLConnectionClientHandler._invoke(URLConnectionClientHandler.java:204) 在 com.sun.jersey.client .urlconnection.URLConnectionClientHandler.handle(URLConnectionClientHandler.java:147) at com.sun.jersey.api.client.Client.handle(Client.java:648) at com.sun.jersey.api.client.WebResource.handle(WebResource .java:670) 在 com.sun.jersey.api.client.WebResource.access$200(WebResource.java:74) 在 com.sun.jersey.api.client.WebResource$Builder.post(WebResource.java:563)在 com.nec.jp.pflow.unc.service.EmployeeService.addEmployee(EmployeeService.java:44) 在 com.nec.jp.pflow.unc.controller.EmployeeController.addCustomer(EmployeeController.java:29) 在 sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method) 在 sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:39) 在 sun.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:25)

But if i convert my json to string representation like :

但是，如果我将我的 json 转换为字符串表示，如：

String input = "{\"name\" : \"" + employee.getName() + "\",\"empId\" : \"" + employee.getEmpId() + "\",\"organizationName\" : \"" + employee.getOrganizationName() + "\"}";

ClientResponse response = service.path("rest").path("vtn").path("addEmplyee")
            .type(MediaType.APPLICATION_JSON_TYPE).post(ClientResponse.class, input);

then it is working fine.

然后它工作正常。

Please suggest how can i put my JSON Object without getting the above exception. What is the best way?

请建议我如何放置我的 JSON 对象而不会出现上述异常。什么是最好的方法？

Thanks in advance.

提前致谢。

---

---

I got a solution for the above. Now I am making use of Hymanson-mapper api for converting the POJO to json.

我得到了上述解决方案。现在我正在使用 Hymanson-mapper api 将 POJO 转换为 json。

Below is the code snippet.

下面是代码片段。

ObjectMapper mapper = new ObjectMapper();
ClientResponse response = resource.path("rest").path("vtn").path("addEmplyee")
        .type(MediaType.APPLICATION_JSON_TYPE).post(ClientResponse.class, mapper.writeValueAsString(employee)); 

回答by Emmanuel Touzery

A better way is to tell jersey-clientthat it may use Hymansonfor serialization: https://stackoverflow.com/a/2580315/516188

更好的方法是告诉jersey-client它可以Hymanson用于序列化：https: //stackoverflow.com/a/2580315/516188

Pasting the setup:

粘贴设置：

ClientConfig cc = new DefaultClientConfig();
cc.getClasses().add(HymansonJsonProvider.class);
Client clientWithHymansonSerializer = Client.create(cc);

HymansonJsonProvideris in the Hymanson-jaxrs-json-providerpackage.

HymansonJsonProvider是在Hymanson-jaxrs-json-provider包中。

回答by Jugal Shah

Just copying the solution from the question above, just in case someone is looking for answer.

只是从上面的问题中复制解决方案，以防万一有人正在寻找答案。

Use Hymanson-mapper api for converting the POJO to json as shown in the code snippet below:

使用 Hymanson-mapper api 将 POJO 转换为 json，如下面的代码片段所示：

ObjectMapper mapper = new ObjectMapper();
ClientResponse response = resource.path("rest").path("vtn").path("addEmplyee")
        .type(MediaType.APPLICATION_JSON_TYPE).post(ClientResponse.class, mapper.writeValueAsString(employee)); 

回答by xparticle

adding a .toString() worked for me:

添加 .toString() 对我有用：

JSONObject object=new JSONObject();
object.put("name", employee.getName());
object.put("empId", employee.getEmpId());
object.put("organizationName", employee.getOrganizationName());


ClientResponse response = service.path("rest").path("vtn").path("addEmplyee").type(MediaType.APPLICATION_JSON_TYPE).post(ClientResponse.class,object.toString());

回答by jaskirat Singh

JSONObject object=new JSONObject();

need to provide @XMLrootElementAnnotation....if you are accessing the web service from remote....i.e XML Mapping..for Model class.

需要提供@XMLrootElementAnnotation....如果您从远程访问 Web 服务....即 XML Mapping..for Model 类。

回答by Suraj Sharma

After lot of reading here is the solution which i arrived at :

经过大量阅读这里是我到达的解决方案：

If you are using JERSEY-CLIENT as specified as below of version 1.9 or higher

如果您使用 JERSEY-CLIENT 如下指定的 1.9 或更高版本

<dependency>
    <groupId>com.sun.jersey</groupId>
    <artifactId>jersey-client</artifactId>
    <version>1.9.1</version>
</dependency>

Then use the MOXY Jar as specified below of version 2.19 or higher

然后使用下面指定的 2.19 或更高版本的 MOXY Jar

<dependency>
    <groupId>org.glassfish.jersey.media</groupId>
    <artifactId>jersey-media-moxy</artifactId>
    <version>2.19</version>
</dependency>

And then add the MOXyJsonProvider.class in the Class of Client configuration as shown below :

然后在 Class of Client 配置中添加 MOXyJsonProvider.class 如下所示：

ClientConfig cc = new DefaultClientConfig();
cc.getClasses().add(MOXyJsonProvider.class);
Client client = Client.create(cc);

Now make the required configuration to Client Object to connect to server and execute the request:

现在对客户端对象进行所需的配置以连接到服务器并执行请求：

WebResource webResource = client.resource(*restFulPath*);
Builder builder = webResource.accept(MediaType.APPLICATION_JSON);

builder.header("Content-Type", "application/json");
builder.header("Accept ", "application/json");
builder.accept(MediaType.APPLICATION_JSON);

try
{
    String requestData = "*requestInJson*";
    ClientResponse response = null;
    response = builder.post(ClientResponse.class, requestData);
    if (response.getStatus() == 200)
    {
        MappedJsonReportOut res = response.getEntity(MappedJsonReportOut.class);
    }
}

回答by abhishek ringsia

Error Which m getting is little Bit Different But Finally Solution is Work for Me

我得到的错误有点不同，但最终解决方案对我有用

   Caused by: com.sun.jersey.api.client.ClientHandlerException: com.sun.jersey.api.client.ClientHandlerException: A message body writer for Java type, class com.siemens.spring.model.User, and MIME media type, application/json, was not found
    at com.sun.jersey.client.urlconnection.URLConnectionClientHandler.handle(URLConnectionClientHandler.java:149)
    at com.sun.jersey.api.client.Client.handle(Client.java:648)
    at com.sun.jersey.api.client.WebResource.handle(WebResource.java:670)
    at com.sun.jersey.api.client.WebResource.access0(WebResource.java:74)
    at com.sun.jersey.api.client.WebResource$Builder.post(WebResource.java:563)
    at com.siemens.controller.UserRegistration.process(UserRegistration.java:156)
    at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method)
    at sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:57)
    at sun.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:43)
    at java.lang.reflect.Method.invoke(Method.java:606)
    at org.apache.el.parser.AstValue.invoke(AstValue.java:278)
    at org.apache.el.MethodExpressionImpl.invoke(MethodExpressionImpl.java:274)
    at javax.faces.event.MethodExpressionActionListener.processAction(MethodExpressionActionListener.java:153)

And This is Solution Worked for Me :

这是对我有用的解决方案：

    Client client = Client.create();
    WebResource webResource = client
            .resource("http://localhost:8080/NewsTickerServices/Siemens/CreateJUser");
    // value added

    ClientResponse response = webResource.type(MediaType.APPLICATION_JSON).post(ClientResponse.class,mapper.writeValueAsString(user));//user is object of User Pojo class 

回答by haggisandchips

You can also use the entity() methods to achieve this, as follows:

您还可以使用 entity() 方法来实现这一点，如下所示：

    ClientResponse response = resource.path("rest").path("vtn").path("addEmplyee")
    .entity(employee, MediaType.APPLICATION_JSON_TYPE)
    .post(ClientResponse.class);

or

或者

    ClientResponse response = resource.path("rest").path("vtn").path("addEmplyee")
    .type(MediaType.APPLICATION_JSON_TYPE)
    .entity(employee)
    .post(ClientResponse.class);