高枕无忧,初始化参数 - 如何访问?
我想在我的 web.xml 中有一些初始化参数,并稍后在应用程序中检索它们,我知道当我有一个普通的 servlet 时我可以做到这一点。然而,使用resteasy,我将HttpServletDispatcher配置为我的默认servlet,所以我不太确定如何从我的休息资源访问它。这可能非常简单,或者我可能需要使用不同的方法,无论哪种方式,了解你们的想法都会很好。以下是我的 web.xml,
<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>RestEasy sample Web Application</display-name>
<!-- <context-param>
<param-name>resteasy.scan</param-name>
<param-value>true</param-value>
</context-param> -->
<listener>
<listener-class>
org.jboss.resteasy.plugins.server.servlet.ResteasyBootstrap
</listener-class>
</listener>
<servlet>
<servlet-name>Resteasy</servlet-name>
<servlet-class>
org.jboss.resteasy.plugins.server.servlet.HttpServletDispatcher
</servlet-class>
<init-param>
<param-name>javax.ws.rs.Application</param-name>
<param-value>com.pravin.sample.YoWorldApplication</param-value>
</init-param>
</servlet>
<servlet-mapping>
<servlet-name>Resteasy</servlet-name>
<url-pattern>/*</url-pattern>
</servlet-mapping>
</web-app>
我的问题是如何在 init-param 中设置某些内容,然后在静态资源中检索它。任何提示将不胜感激。谢谢你们!
I'd like to have some init params in my web.xml and retrieve them later in the application, I know I can do this when I have a normal servlet. However with resteasy I configure HttpServletDispatcher to be my default servlet so I'm not quite sure how I can access this from my rest resource. This might be completely simple or I might need to use a different approach, either way it would be good to know what you guys think. Following is my web.xml,
<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>RestEasy sample Web Application</display-name>
<!-- <context-param>
<param-name>resteasy.scan</param-name>
<param-value>true</param-value>
</context-param> -->
<listener>
<listener-class>
org.jboss.resteasy.plugins.server.servlet.ResteasyBootstrap
</listener-class>
</listener>
<servlet>
<servlet-name>Resteasy</servlet-name>
<servlet-class>
org.jboss.resteasy.plugins.server.servlet.HttpServletDispatcher
</servlet-class>
<init-param>
<param-name>javax.ws.rs.Application</param-name>
<param-value>com.pravin.sample.YoWorldApplication</param-value>
</init-param>
</servlet>
<servlet-mapping>
<servlet-name>Resteasy</servlet-name>
<url-pattern>/*</url-pattern>
</servlet-mapping>
</web-app>
My question is how do I set something in the init-param and then retrieve it later in a restful resource. Any hints would be appreciated. Thanks guys!
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使用 @Context 注释将您想要的任何内容注入到您的方法中:
使用 @Context,您可以注入 HttpHeaders、UriInfo、Request、HttpServletRequest、HttpServletResponse、ServletConvig、ServletContext、SecurityContext。
或者如果您使用此代码,则可以执行其他任何操作:
Use the @Context annotation to inject whatever you want into your method:
With @Context you can inject HttpHeaders, UriInfo, Request, HttpServletRequest, HttpServletResponse, ServletConvig, ServletContext, SecurityContext.
Or anything else if you use this code: