以编程方式配置Jersey应用程序

Question

我正在尝试使用我自己的应用程序实现或扩展的ResourceConfig或PackageResourceConfig来配置我的Jersey应用程序。因此，我的第一次尝试是将现有的web.xml（实际上，由于开发的库性质，我使用的是web-fragment.xml）配置移植到MyApplication实现。

<servlet>
    <servlet-name>rest</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>my.pkg.resources;org.codehaus.jackson.jaxrs</param-value>
    </init-param>
    <init-param>
        <param-name>com.sun.jersey.api.json.POJOMappingFeature</param-name>
        <param-value>true</param-value>
    </init-param>
    <init-param>
        <param-name>com.sun.jersey.spi.container.ContainerRequestFilters</param-name>
        <param-value>my.pkg.Filter1;my.pkg.Filter2</param-value>
    </init-param>
</servlet>
<servlet-mapping>
    <servlet-name>rest</servlet-name>
    <url-pattern>/rest/*</url-pattern>
</servlet-mapping>

<servlet>
    <servlet-name>my.pkg.MyApplication</servlet-name> <!-- implementation follows -->
    <init-param>
        <param-name>com.sun.jersey.api.json.POJOMappingFeature</param-name>
        <param-value>true</param-value>
    </init-param>
</servlet>
<servlet-mapping>
    <servlet-name>my.pkg.MyApplication</servlet-name>
    <url-pattern>/rest/*</url-pattern>
</servlet-mapping>

// [...]

public class MyApplication extends PackagesResourceConfig {
private static final Logger logger = Logger.getLogger(MyApplication.class);

@Context
ServletConfig config

public MyApplication() {

    super("my.pkg.resources;org.codehaus.jackson.jaxrs");

    super.getFeatures().put(JSONConfiguration.FEATURE_POJO_MAPPING, Boolean.TRUE);

    // filters not set
}

@PostConstruct 
public void readInitParams() {

    // read init params from ServletConfig

    // config.getInitParameterNames();
    // ...
}

}

当我使用第二个版本时，我会收到以下信息

mag 27, 2013 12:08:03 PM com.sun.jersey.api.core.PackagesResourceConfig init
INFO: Scanning for root resource and provider classes in the packages:
  aero.aice.jerico.rs;org.codehaus.jackson.jaxrs;
mag 27, 2013 12:08:03 PM com.sun.jersey.server.impl.application.DeferredResourceConfig$ApplicationHolder <init>
INFO: Instantiated the Application class my.package.MyApplication. The following root resource and provider classes are registered: [class com.sun.jersey.server.impl.application.WebApplicationImpl$1WadlContextResolver, class org.codehaus.jackson.jaxrs.JsonParseExceptionMapper, class org.codehaus.jackson.jaxrs.JacksonJaxbJsonProvider, class org.codehaus.jackson.jaxrs.JsonMappingExceptionMapper, class aero.aice.jerico.rs.service.OperationService, class aero.aice.jerico.rs.service.CrudService, class org.codehaus.jackson.jaxrs.JacksonJsonProvider]
mag 27, 2013 12:08:03 PM com.sun.jersey.core.spi.component.ProviderFactory __getComponentProvider
SEVERE: The provider class, class com.sun.jersey.server.impl.application.WebApplicationImpl$1WadlContextResolver, could not be instantiated. Processing will continue but the class will not be utilized
java.lang.InstantiationException: com.sun.jersey.server.impl.application.WebApplicationImpl$1WadlContextResolver

mag 27, 2013 12:08:04 PM com.sun.jersey.spi.inject.Errors processErrorMessages
SEVERE: The following errors and warnings have been detected with resource and/or provider classes:
  SEVERE: The class com.sun.jersey.server.impl.application.WebApplicationImpl$1WadlContextResolver is a not a public class and cannot be instantiated.
  SEVERE: The inner class com.sun.jersey.server.impl.application.WebApplicationImpl$1WadlContextResolver is not a static inner class and cannot be instantiated.

正如您所看到的，com.sun.jersey.server.impl.application.WebApplicationImpl$1WADLContextResolver是第一个注册的类，但由于它不是公共的，也不是静态的内部类，所以它是不可实例化的。我发现关于这个过程的文档很少，特别是关于如何在初始化期间设置特性和属性的文档。我使用的是Jersey的最新版本(1.17.1)，但也用1.9进行了测试。是否也可以通过编程设置应用程序路径？我在文档中看到过@ApplicationPath,但它对我的目的并不有用,因为我需要在运行时设置它。我知道这些是更多的问题，但我认为它们都有相同的根源。谁能给我指出正确的方向吗？

Answer 1

确保在扫描资源类的路径上没有jersey库。

只能在部署期间设置的基URI。使用war名称或@applicationpath。您可以使用web.xml中的servlet-mapping元素重写它。