【发布时间】:2018-05-17 01:07:48
【问题描述】:
我的应用程序中有两个 servlet,我想显示它们的内容,所以我这样做了:
<servlet>
<servlet-name>DisciplinaService</servlet-name>
<servlet-class>org.glassfish.jersey.servlet.ServletContainer</servlet-class>
<init-param>
<param-name>javax.ws.rs.Application</param-name>
<param-value>com.lab4.club.main.MiApp</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet>
<servlet-name>SocioService</servlet-name>
<servlet-class>org.glassfish.jersey.servlet.ServletContainer</servlet-class>
<init-param>
<param-name>javax.ws.rs.Application</param-name>
<param-value>com.lab4.club.main.MiApp</param-value>
</init-param>
<load-on-startup>2</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>DisciplinaService</servlet-name>
<url-pattern>/*</url-pattern>
</servlet-mapping>
<servlet-mapping>
<servlet-name>SocioService</servlet-name>
<url-pattern>/*</url-pattern>
</servlet-mapping>
当我输入“../disciplinas”或“../socios”时,出现 404 错误。 但是,如果我只有一个映射的 servlet,像这样:
<servlet>
<servlet-name>DisciplinaService</servlet-name>
<servlet-class>org.glassfish.jersey.servlet.ServletContainer</servlet-class>
<init-param>
<param-name>javax.ws.rs.Application</param-name>
<param-value>com.lab4.club.main.MiApp</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>DisciplinaService</servlet-name>
<url-pattern>/*</url-pattern>
</servlet-mapping>
没有问题。
谁能帮我展示这两个 servlet?问题出在 web.xml 中,我不知道两个如何显示多个 servlet。请注意,我没有使用 spring 或 maven。
【问题讨论】:
标签: java spring maven servlets jersey