多个平针织servlet在一个web.xml中

时间：2012-07-12 21:29:20

标签： java spring rest jersey

是否可以在一个web.xml中包含多个jersey servlet？我试图以这种方式进行RESTfull版本控制：

<?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>myapi</display-name>

  <context-param>
    <param-name>contextConfigLocation</param-name>
    <param-value>/WEB-INF/context-v1.xml /WEB-INF/context-v2.xml</param-value>
  </context-param>

  <listener>
    <listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
  </listener>

  <servlet>
    <servlet-name>REST-V1</servlet-name>
    <servlet-class>com.sun.jersey.spi.spring.container.servlet.SpringServlet</servlet-class>
    <init-param>
      <param-name>com.sun.jersey.config.property.packages</param-name>
      <param-value>com.myapi.rest.v1</param-value>
    </init-param>
    <load-on-startup>1</load-on-startup>
  </servlet>

  <servlet-mapping>
    <servlet-name>REST-V1</servlet-name>
    <url-pattern>/v1/*</url-pattern>
  </servlet-mapping>

  <servlet>
    <servlet-name>REST-V2</servlet-name>
    <servlet-class>com.sun.jersey.spi.spring.container.servlet.SpringServlet</servlet-class>
    <init-param>
      <param-name>com.sun.jersey.config.property.packages</param-name>
      <param-value>com.myapi.rest.v2</param-value>
    </init-param>
    <load-on-startup>2</load-on-startup>
  </servlet>

  <servlet-mapping>
    <servlet-name>REST-V2</servlet-name>
    <url-pattern>/v2/*</url-pattern>
  </servlet-mapping>

  <welcome-file-list>
    <welcome-file>index.html</welcome-file>
  </welcome-file-list>

</web-app>

但应该单独加载spring context-v1和context-v2？因为他们有豆子，它们具有相同的名称等。

修改

如果查看我的控制台输出，它会为每个servlet加载两次资源（admin / info）：

15.07.2012 14:47:08 com.sun.jersey.api.core.PackagesResourceConfig init
INFO: Scanning for root resource and provider classes in the packages:
  com.myapi.rest.v1
15.07.2012 14:47:08 com.sun.jersey.api.core.ScanningResourceConfig logClasses
INFO: Root resource classes found:
  class com.myapi.rest.v1.LOAdminResource
  class com.myapi.rest.v1.LOInfoResource
15.07.2012 14:47:08 com.sun.jersey.api.core.ScanningResourceConfig init
INFO: No provider classes found.
15.07.2012 14:47:09 com.sun.jersey.spi.spring.container.servlet.SpringServlet getContext
INFO: Using default applicationContext
15.07.2012 14:47:09 com.sun.jersey.spi.spring.container.SpringComponentProviderFactory registerSpringBeans
INFO: Registering Spring bean, adminResource_v2, of type com.myapi.rest.v2.LOAdminResource as a root resource class
15.07.2012 14:47:09 com.sun.jersey.spi.spring.container.SpringComponentProviderFactory registerSpringBeans
INFO: Registering Spring bean, infoResource_v2, of type com.myapi.rest.v2.LOInfoResource as a root resource class
15.07.2012 14:47:09 com.sun.jersey.spi.spring.container.SpringComponentProviderFactory registerSpringBeans
INFO: Registering Spring bean, adminResource_v1, of type com.myapi.rest.v1.LOAdminResource as a root resource class
15.07.2012 14:47:09 com.sun.jersey.spi.spring.container.SpringComponentProviderFactory registerSpringBeans
INFO: Registering Spring bean, infoResource_v1, of type com.myapi.rest.v1.LOInfoResource as a root resource class
15.07.2012 14:47:09 com.sun.jersey.server.impl.application.WebApplicationImpl _initiate
INFO: Initiating Jersey application, version 'Jersey: 1.8 06/24/2011 12:17 PM'
15.07.2012 14:47:09 com.sun.jersey.api.core.PackagesResourceConfig init
INFO: Scanning for root resource and provider classes in the packages:
  com.myapi.rest.v2
15.07.2012 14:47:09 com.sun.jersey.api.core.ScanningResourceConfig logClasses
INFO: Root resource classes found:
  class com.myapi.rest.v2.LOAdminResource
  class com.myapi.rest.v2.LOInfoResource
15.07.2012 14:47:09 com.sun.jersey.api.core.ScanningResourceConfig init
INFO: No provider classes found.
15.07.2012 14:47:09 com.sun.jersey.spi.spring.container.servlet.SpringServlet getContext
INFO: Using default applicationContext
15.07.2012 14:47:09 com.sun.jersey.spi.spring.container.SpringComponentProviderFactory registerSpringBeans
INFO: Registering Spring bean, adminResource_v2, of type com.myapi.rest.v2.LOAdminResource as a root resource class
15.07.2012 14:47:09 com.sun.jersey.spi.spring.container.SpringComponentProviderFactory registerSpringBeans
INFO: Registering Spring bean, infoResource_v2, of type com.myapi.rest.v2.LOInfoResource as a root resource class
15.07.2012 14:47:09 com.sun.jersey.spi.spring.container.SpringComponentProviderFactory registerSpringBeans
INFO: Registering Spring bean, adminResource_v1, of type com.myapi.rest.v1.LOAdminResource as a root resource class
15.07.2012 14:47:09 com.sun.jersey.spi.spring.container.SpringComponentProviderFactory registerSpringBeans
INFO: Registering Spring bean, infoResource_v1, of type com.myapi.rest.v1.LOInfoResource as a root resource class
15.07.2012 14:47:09 com.sun.jersey.server.impl.application.WebApplicationImpl _initiate
INFO: Initiating Jersey application, version 'Jersey: 1.8 06/24/2011 12:17 PM'

4 个答案:

答案 0 :(得分：8)

是的，您可以在web.xml中指定两个或更多servlet。请记住为每个servlet映射指定不同的servlet映射。

<servlet>
    <servlet-name>servletOne</servlet-name>
    <servlet-class>com.sun.jersey.spi.spring.container.servlet.SpringServlet</servlet-class>
    <init-param>
    <param-name>com.sun.jersey.config.property.packages</param-name>
    <param-value>com.packageOne</param-value>
    </init-param>
    <load-on-startup>1</load-on-startup>
</servlet>
<servlet>
    <servlet-name>servletTwo</servlet-name>
    <servlet-class>com.sun.jersey.spi.spring.container.servlet.SpringServlet</servlet-class>
    <init-param>
    <param-name>com.sun.jersey.config.property.packages</param-name>
    <param-value>com.packageTwo</param-value>
    </init-param>
    <load-on-startup>2</load-on-startup>
</servlet>

<servlet-mapping>
    <servlet-name>ServletOne</servlet-name>
    <url-pattern>/v1/*</url-pattern>
</servlet-mapping>
    <servlet-mapping>
    <servlet-name>ServletTwo</servlet-name>
<url-pattern>/v2/*</url-pattern>
</servlet-mapping>

initParameter loadOnStartup定义了加载servlet的顺序（在本例中是第一个servletOne，然后是servletTwo）。

答案 1 :(得分：6)

当你一起使用Jersey和Spring时，Jersey / Spring servlet遍历所有可用的Spring bean并注册它们将在其中找到的每个资源和提供者类。

如果有多个Jersey / Spring servlet使用相同的（根）上下文并因此共享bean定义，那么将为每个这样的servlet执行该过程，并且多次注册资源和提供者类。

为了避免多次注册同一个bean，在相应的Jesrey / Spring servlet的子上下文中定义这样的bean。

除非需要混合使用Spring-managed和Jersey-managed类，否则甚至没有必要提供用于在web.xml中声明类的初始化参数。

<context-param>
    <param-name>contextConfigLocation</param-name>
    <param-value>/WEB-INF/root-context.xml</param-value>
</context-param>

...

<servlet>
    <servlet-name>REST-V1</servlet-name>
    <servlet-class>com.sun.jersey.spi.spring.container.servlet.SpringServlet</servlet-class>
    <init-param>
        <param-name>contextConfigLocation</param-name>
        <param-value>/WEB-INF/context-v1.xml</param-value>
    </init-param>
    <load-on-startup>1</load-on-startup>
</servlet>

<servlet>
    <servlet-name>REST-V2</servlet-name>
    <servlet-class>com.sun.jersey.spi.spring.container.servlet.SpringServlet</servlet-class>
    <init-param>
        <param-name>contextConfigLocation</param-name>
        <param-value>/WEB-INF/context-v1.xml /WEB-INF/context-v2.xml</param-value>
    </init-param>
    <load-on-startup>2</load-on-startup>
</servlet>

...

答案 2 :(得分：1)

我知道这个话题很难回答。但我的回答可以帮助别人。

我们可以使用这些分隔符在web.xml中配置多个资源包：

  1. 白色空间
  2. 逗号（，）
  3. 分号（;）
  4. Nextline

    5. 示例： 

    <?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_3_0.xsd" id="WebApp_ID" version="3.0">
  <display-name>MultiplePackageRest</display-name>
  <servlet>
    <servlet-name>JerseyMultiplePackage</servlet-name>
    <servlet-class>org.glassfish.jersey.servlet.ServletContainer</servlet-class>
    <init-param>
      <param-name>jersey.config.server.provider.packages</param-name>
      <param-value>info.javadoff.rest1,info.javadoff.rest2,...</param-value>
    </init-param>
      <load-on-startup>1</load-on-startup>
  </servlet>
  <servlet-mapping>
    <servlet-name>JerseyMultiplePackage</servlet-name>
    <url-pattern>/api/*</url-pattern>
  </servlet-mapping>
</web-app>

答案 3 :(得分：1)

另一种可能性是覆盖com.sun.jersey.spi.spring.container.servlet.SpringServlet中的一个方法。 initiate方法看起来像这样（版本1.19.1）：

@Override
protected void initiate(ResourceConfig rc, WebApplication wa) {
    try {
        wa.initiate(rc, new SpringComponentProviderFactory(rc, getContext()));
    } catch (RuntimeException e) {
        LOGGER.log(Level.SEVERE, "Exception occurred when intialization", e);
        throw e;
    }
}

如果您更改这样的子类中的代码，那么您可以根据您的条件（例如包名称）过滤掉不需要的spring bean：

@Override
protected void initiate(ResourceConfig rc, WebApplication wa) {
    try {
        SpringComponentProviderFactory springComponentProviderFactory = new SpringComponentProviderFactory(rc, getContext());
        rc.getClasses().removeIf( clazz -> clazz.getPackage().getName().startsWith( "bla" ));
        wa.initiate(rc, springComponentProviderFactory);
    } catch (RuntimeException e) {
        LOGGER.log(Level.SEVERE, "Exception occurred when intialization", e);
        throw e;
    }
}

有点hacky解决方案，但对我们来说非常有效。

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