Question

我在网上看到很多人这样做的例子......

<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns="http://java.sun.com/xml/ns/javaee"
         xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
         xsi:schemaLocation="http://java.sun.com/xml/ns/javaee
http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd"
     version="3.0">
<servlet-mapping>
<servlet-name>javax.ws.rs.core.Application</servlet-name>
<url-pattern>/hello/*</url-pattern>
</servlet-mapping>
</web-app>

然而，这对我来说无效。我收到错误

java.lang.IllegalArgumentException: Servlet mapping specifies an unknown servlet name javax.ws.rs.core.Application

那么能够直接在servlet-name中使用类名的秘诀是什么？

Answer 1

我认为@davideconsonni几乎是对的：

<servlet>
    <servlet-name>HelloServlet</servlet-name>
    <servlet-class>javax.ws.rs.core.Application</servlet-class>
    <load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
    <servlet-name>HelloServlet</servlet-name>
    <url-pattern>/hello/*</url-pattern>
</servlet-mapping>

Answer 2

首先声明你的servlet：

<servlet>
    <servlet-name>javax.ws.rs.core.Application</servlet-name>
    <load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
    <servlet-name>javax.ws.rs.core.Application</servlet-name>
    <url-pattern>/hello/*</url-pattern>
</servlet-mapping>

Web.xml <servlet-name>可以只是一个类吗？</servlet-name>

2 个答案: