maven web.xml元素类型“web-app”必须以匹配的结束标记“”结束

Question

Eclipse Java EE IDE
Apache Tomcat V9.0
我通过new->Maven project创建了项目，并选择Maven原型作为maven-archetype-webapp。
web.xml内容如下所示。

<?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"
    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>WebProject</display-name>
    <servlet>
        <servlet-name>LoginServlet</servlet-name>
        <display-name>LoginServlet</display-name>
        <servlet-class>com.wise.LoginServlet</servlet-class>
    </servlet>
    <servlet-mapping>
        <servlet-name>LoginServlet</servlet-name>
        <url-pattern>/LoginServlet</url-pattern>
    </servlet-mapping>
</web-app>

eclipse中的项目结构如下所示

Answer 1

您的web.xml结构应该如下所示。

<?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"
    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>WebProject</display-name>
    <servlet>
        <servlet-name>LoginServlet</servlet-name>
        <display-name>LoginServlet</display-name>
        <servlet-class>com.wise.LoginServlet</servlet-class>
    </servlet>
    <servlet-mapping>
        <servlet-name>LoginServlet</servlet-name>
        <url-pattern>/LoginServlet</url-pattern>
    </servlet-mapping>
</web-app>