使用Jersey的JAXRS Rest服务中的问题 [英] Issues in JAXRS Rest service using jersey

问题描述

使用球衣JAXRS API在tomcat 7上部署了一个演示Rest服务示例.开发了资源类

Deployed a demo Rest Service on tomcat 7 using jersey JAXRS API . Developed a resource class

@Path("/supportdata")
public class SupportDataService {
    public SupportDataService() {
        // TODO Auto-generated constructor stub
    }


    @GET
    @Produces(MediaType.APPLICATION_XML)
    public String getSupportData(){
        String xmlSupport=null;

        xmlSupport="<SupportData><Support><key>path1</key><value>value1</value></Support><Support><key>path2</key><value>value2</value></Support></SupportData>";

        return xmlSupport;
    }
}

应用程序的子类

public class RestApplication extends Application {
    public RestApplication() {
        // TODO Auto-generated constructor stub
    }

    @Override
    public Set<Class<?>> getClasses() {
        // TODO Auto-generated method stub

        Set<Class<?>> s=new HashSet<Class<?>>();
        s.add(SupportDataService.class);
        return s;
    }
}
 Web.xml contains

<web-app>
 <servlet>
        <servlet-name>com.jaxrs.RestApplication</servlet-name>
    </servlet>
    <servlet-mapping>
        <servlet-name>com.jaxrs.RestApplication</servlet-name>
        <url-pattern>/resources</url-pattern>
    </servlet-mapping>
</web-app>

Included javax.ws.rs-api-2.0.1.jar from jersey in the project. Deployed it on defaultport of tomcat 7 and calling the url  on local host port 8080 and url 

/RestService/resources/supportdata

/RestService/resources/supportdata

获取404错误的资源不可用.

getting a 404 error of resource not available.

推荐答案

给出...

1. RestApplication在com.jaxrs软件包中

您具有必需的依赖项 ^[1]

You have the required dependencies ^[1]

没有其他未知问题与您未向我们展示的内容有关

No other unknown issues related to something you're not showing us

您需要做的所有工作才能使它起作用...

正在改变

All you need to do to get this to work...

is change

 <url-pattern>/resources</url-pattern>

到

 <url-pattern>/resources/*</url-pattern>

应该可以通过http://localhost:8080/yourapp/resources/supportdata

对此进行了测试，效果很好. /resources将网址格式严格限制为/resources.添加/*时，您会说带有/resources前缀的任何内容.

Tested this and it works fine. /resources is limiting the url pattern to strictly /resources. When you add the /* you are saying anything with the /resources prefix.

[1] :

使用Maven，这是我使用的唯一依赖项

Using Maven, this is the only dependency I used

<dependency>
    <groupId>org.glassfish.jersey.containers</groupId>
    <artifactId>jersey-container-servlet</artifactId>
    <version>2.13</version>
</dependency>

不使用Maven(啊，为什么?):

Not using Maven (ahhhh, whyyy?):

我建议您从此处下载 RI包，并将所有jar包含到您的项目中.

I suggest you download the RI Bundle from here and include all the jars into your project.

只是为了让其他人为OP的配置scratch之以鼻，这来自JAX-RS规范:

And just for others scratching their head about the OP's configuration, this is from the JAX-RS spec:

如果存在Application子类:

• 如果已经有一个处理该应用程序的servlet.也就是说，一个Servlet的初始化参数为

• If there is already a servlet that handles this application. That is, a servlet that has an initialization parameter named

       javax.ws.rs.core.Application

       javax.ws.rs.core.Application

其值是Application子类的全限定名，那么JAX-RS实现无需任何其他配置步骤.

whose value is the fully qualified name of the Application subclass, then no additional configuration steps are required by the JAX-RS implementation.

如果没有servlet处理此应用程序，则需要JAX-RS实现来动态添加其完全限定名称必须是Application子类名称的servlet.如果Application子类用@ApplicationPath注释，则要求实现使用此注释的值附加"/*"来定义所添加服务器的映射.否则，必须将应用程序与指定servlet映射的web.xml打包在一起.例如，如果org.example.MyApplication是Application子类的名称，则示例web.xml将是:

If no servlet handles this application, JAX-RS implementations are REQUIRED to dynamically add a servlet whose fully qualified name must be that of the Application subclass. If the Application subclass is annotated with @ApplicationPath, implementations are REQUIRED to use the value of this annotation appended with "/*" to define a mapping for the added server. Otherwise, the application MUST be packaged with a web.xml that specifies a servlet mapping. For example, if org.example.MyApplication is the name of the Application subclass, a sample web.xml would be:

1   <web-app version="3.0" xmlns="http://java.sun.com/xml/ns/javaee"
2                        xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
3                        xsi:schemaLocation="http://java.sun.com/xml/ns/javaee
4                        http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd">
5       <servlet>
6            <servlet-name>org.example.MyApplication</servlet-name>
7       </servlet>
8       <servlet-mapping>
9            <servlet-name>org.example.MyApplication</servlet-name>
10           <url-pattern>/myresources/*</url-pattern>
11      </servlet-mapping>
12  </web-app>

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