一个 web.xml 中的多个 jersey servlet答案

【问题标题】：Multiple jersey servlets in one single web.xml一个 web.xml 中的多个 jersey servlet
【发布时间】：2012-07-12 21:29:20
【问题描述】：

是否可以在一个 web.xml 中包含多个 jersey servlet？我正在尝试以这种方式进行 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>

但是spring context-v1和context-v2应该分开加载？因为他们有豆子，它们有相同的名字等等。

编辑：

如果您查看我的控制台输出，它会为每个 servlet 加载资源（管理员/信息）两次：

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'

【问题讨论】：

load-on-startup 应该——如果我没记错的话——在这两种情况下都是 1
不，这不是真的，它是 servlet 的启动顺序...

标签： java spring rest jersey

【解决方案1】：

是的，您可以在 web.xml 中指定两个或多个 servlet。请记住为每一个指定不同的 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（在这种情况下，首先是 servletOne，然后是 servletTwo）。

【讨论】：

没错，两个 servlet 都已加载。但请查看我的控制台输出。我的问题是我的资源（AdminResource、InfoResource）为每个 servlet 加载了两次...
mmm... 很奇怪，因为您已经定义了两个不同的包... 或者您可以尝试以编程方式定义服务器和 Servlet（如此处定义的代码 stackoverflow.com/questions/11210734/… 在 Start Server Android Activity 下)
这不是正确答案！无论您放什么，Spring 都会加载它在主上下文中找到的每个带注释的类。因此，您将拥有两个提供完全相同的东西的 servlet。当心并尝试一下！正确答案在上面……

【解决方案2】：

问题是，当您同时使用 Jersey 和 Spring 时，Jersey/Spring servlet 会遍历所有可用的 Spring bean 并注册它将在其中找到的每个资源和提供程序类。

如果您有多个 Jersey/Spring servlet 使用相同的（根）上下文并因此共享 bean 定义，则为每个此类 servlet 执行该过程，并且资源和提供者类被注册多次。

为了避免多次注册同一个 bean，请在相应 Jesrey/Spring servlet 的子上下文中定义此类 bean。

甚至不需要为在 web.xml 中声明类提供初始化参数，除非需要混合使用 Spring 管理和 Jersey 管理的类。

<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>

...

【讨论】：

这应该是选择的答案。

【解决方案3】：

我知道这个话题已经过时了。但我的回答可以帮助别人。

我们可以用这些分隔符在web.xml中配置多个资源包：

空白
逗号 (,)
分号 (;)
下一行

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>

【讨论】：

【解决方案4】：

另一种可能性是覆盖com.sun.jersey.spi.spring.container.servlet.SpringServlet 中的一种方法。启动方法如下所示（1.19.1 版）：

@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;
    }
}

如果您像这样更改子类中的代码，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;
    }
}

有点老套的解决方案，但对我们非常有效。

【讨论】：