打开jsp界面为代码

Question

我把xml文件servlt注释了就可以到正常，但是我要用xml的servlet这种应该怎么办呢
没注释前，他的路径会变为file这种，怎么让她变回http://localhost:8080/book_war/pages/user/regist.jsp这种路径
不懂，求支个招。

<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns="http://xmlns.jcp.org/xml/ns/javaee"
         xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
         xsi:schemaLocation="http://xmlns.jcp.org/xml/ns/javaee http://xmlns.jcp.org/xml/ns/javaee/web-app_4_0.xsd"
         version="4.0">
    <servlet>
        <servlet-name>RegistServlet</servlet-name>
        <servlet-class>com.example.book.web.RegistServlet</servlet-class>
    </servlet>

    <servlet-mapping>
        <servlet-name>RegistServlet</servlet-name>
        <url-pattern>/registServlet</url-pattern>
    </servlet-mapping>

    <servlet>
        <servlet-name>LoginServlet</servlet-name>
        <servlet-class>com.example.book.web.LoginServlet</servlet-class>
    </servlet>

    <servlet-mapping>
        <servlet-name>LoginServlet</servlet-name>
        <url-pattern>/loginServlet</url-pattern>
    </servlet-mapping>
</web-app>

Answer 1

https://blog.csdn.net/crayon0/article/details/122272046?spm=1001.2101.3001.6650.5&utm_medium=distribute.pc_relevant.none-task-blog-2%7Edefault%7EBlogCommendFromBaidu%7Edefault-5-122272046-blog-113795918.pc_relevant_default&depth_1-utm_source=distribute.pc_relevant.none-task-blog-2%7Edefault%7EBlogCommendFromBaidu%7Edefault-5-122272046-blog-113795918.pc_relevant_default&utm_relevant_index=9