Question

服务请求类工作正常，服务请求助手也一样，但是userauthservice没有被命中！程序退出时发现404未找到。

这是我的ServiceRequest类

public class ServiceFilter implements ContainerRequestFilter{


@Context
HttpServletRequest servletRequest;
@Override
public ContainerRequest filter(ContainerRequest request) {

    System.out.println("request : "+request);
    URI reqUri = request.getRequestUri();
    System.out.println("req uri "+reqUri );
    BufferedReader inFromClient = null;

    try {

        inFromClient = new BufferedReader(new InputStreamReader(
                request.getEntityInputStream(), "UTF-8"));

    } catch (UnsupportedEncodingException e1) {

        e1.printStackTrace();

    }

    String line;

    StringBuffer buff = new StringBuffer();

    try {

        while ((line = inFromClient.readLine()) != null) {

            buff.append(line);

        }

        inFromClient.close();

    } catch (IOException e) {

        e.printStackTrace();

    }

    System.out.println("###### Request parameters- " + buff);

    String userName = null;

    String password = null;

    Long sessionId = null;

    String serviceMethodName = request.getPathSegments()
            .get(request.getPathSegments().size() - 1).getPath();

    if (!NullObjects.isNull(reqUri)
            && !reqUri.toString().contains(
                    CommonConstants.REPORT_SERVICE_URI)) {

        Response response = null;
        try {
            System.out.println("service method name : "+serviceMethodName);
            ObjectMapper mapper = new ObjectMapper();

            Map filterParams = mapper.readValue(buff.toString(),HashMap.class);
            //call helper
            ServiceRequestFilterHelper helper = new ServiceRequestFilterHelper();
            request = helper.setRequestParams(request, filterParams, serviceMethodName);

        } catch (Exception e) {
            e.printStackTrace();
            throw new WebApplicationException(response);
        }
    }
    System.out.println("request"+request);
    System.out.println("request URI: "+request.getRequestUri());
    System.out.println("request-QueryParameters-"+request.getQueryParameters());
    return request;
    }
}

这是我的serviceRequestFilterHelper Class

public class ServiceRequestFilterHelper {

String userName = null;
String password = null;

public ContainerRequest setRequestParams(ContainerRequest request,  Map reqParam, String serviceMethodName) {
    if (serviceMethodName.equalsIgnoreCase(CommonConstants.SERVICE_METHOD_VALIDATE_CREDENTAILS)) {
        request = setRequestParamValidateCreditials(request, reqParam);

    }
    return request;
}

这是我的UserService类

@Path("/userauthservice")
@PerSession
public class UserService implements Serializable{


    @Path("/validateCredentails")
    @POST
    @Produces("application/json")
    public Response validateCredentails(@QueryParam("user") String user,
                                    @QueryParam("password") String password)
                        throws JSONException{
        LoginDao loginDao = new LoginDao();
        List<TAccessAuthorization> result = loginDao.list(user, password);
        System.out.println("result : "+result.get(0).getAaCode());
        return Response.status(200).entity(result).build();
    }

}

这是我的web.xml

<?xml version="1.0" encoding="UTF-8"?>  
<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"     xmlns="http://java.sun.com/xml/ns/javaee" xmlns:web="http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd" id="WebApp_ID" version="2.5">  
    <display-name>RESTfulWebServiceExample</display-name>  
    <servlet>  
        <servlet-name>Jersey REST Service</servlet-name>  
        <servlet-class>com.sun.jersey.spi.container.servlet.ServletContainer</servlet-class>  
        <init-param>  
            <param-name>com.sun.jersey.config.property.packages</param-name>  
            <param-value>org.arpit.javapostsforlearning.webservice</param-value>  
        </init-param>
        <init-param>
            <param-name>com.sun.jersey.spi.container.ContainerRequestFilters</param-name>
            <param-value>com.easyskool.filter.ServiceFilter</param-value>
        </init-param>  
        <load-on-startup>1</load-on-startup>  
    </servlet>  
    <servlet-mapping>  
        <servlet-name>Jersey REST Service</servlet-name>  
        <url-pattern>/rest/*</url-pattern>  
    </servlet-mapping>  
</web-app>

这是我在XHR海报上看到的输出 POST on // localhost：8080 / test / rest / userauthservice / validateCredentails 状态：404未找到

输入值

网址： - // localhost：8080 / test / rest / userauthservice / validateCredentails

邮寄正文： - {“user”：“webadmin”，“password”：“webadmin”}

Answer 1

XML文件中存在更新的xml低于

的问题

   <?xml version="1.0" encoding="UTF-8"?>  
  <web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"        xmlns="http://java.sun.com/xml/ns/javaee" xmlns:web="http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee   http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd" id="WebApp_ID" version="2.5">  
<display-name>RESTfulWebServiceExample</display-name>  
<servlet>  
    <servlet-name>Jersey REST Service</servlet-name>  
    <servlet-class>com.sun.jersey.spi.container.servlet.ServletContainer</servlet-class>  
    <init-param>
        <param-name>com.sun.jersey.spi.container.ContainerRequestFilters</param-name>
        <param-value>com.easyskool.filter.ServiceFilter</param-value>
    </init-param>  
    <load-on-startup>1</load-on-startup>  
</servlet>  
<servlet-mapping>  
    <servlet-name>Jersey REST Service</servlet-name>  
    <url-pattern>/rest/*</url-pattern>  
</servlet-mapping>

通过containerRequest后发布请求显示404未找到

1 个答案: