Question

我试图通过使用WEB-INF目录中web.xml文件中的配置来呈现para值，但我只是The default user from the servlet config is:呈现（没有web.xml文件的parm值））。有人可以解释我的错误在哪里吗？

initpage.jps

<%@ page language="java" contentType="text/html; charset=ISO-8859-1"
    pageEncoding="ISO-8859-1"%>
<!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd">
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=ISO-8859-1">
<title>Insert title here</title>
</head>
<body>
The default user from the servlet config is:
<%getServletConfig().getInitParameter("defaultUser");%>

</body>
</html>

的web.xml

<?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>SimpleServletProject</display-name>
    <welcome-file-list>
        <welcome-file>index.html</welcome-file>
    </welcome-file-list>

    <servlet>
        <servlet-name>InitJSP</servlet-name>
        <jsp-file>/initpage.jsp</jsp-file>
        <init-param>
        <param-name>defaultUser</param-name>
        <param-value>Default User Name</param-value>
        </init-param>
    </servlet>
    <servlet-mapping>
      <servlet-name>InitJSP</servlet-name>
      <url-pattern>initpage.jsp</url-pattern>
    </servlet-mapping>

</web-app>

在initpage.jps文件中呈现web.xml的param-value

0 个答案: