我正在尝试通过web.xml文件将一些init-param值加载到serevlet中,但它们一直显示为null。我确实有两个上下文参数可以正常工作。有谁知道我做错了什么?
这是我用于我的应用程序的web.xml文件。
<?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_3_1.xsd"
version="3.1"
>
<display-name>Hello World Application</display-name>
<servlet>
<servlet-name>contextxParameterServlet</servlet-name>
<servlet-class>com.wrox.HelloServlet</servlet-class>
<init-param>
<param-name>database</param-name>
<param-value>CustomerSupport</param-value>
</init-param>
<init-param>
<param-name>server</param-name>
<param-value>10.0.12.5</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>contextxParameterServlet</servlet-name>
<url-pattern>/contextParameters</url-pattern>
</servlet-mapping>
<context-param>
<param-name>settingOne</param-name>
<param-value>foo</param-value>
</context-param>
<context-param>
<param-name>settingTwo</param-name>
<param-value>bar</param-value>
</context-param>
<servlet>
<servlet-name>servletParameterServlet</servlet-name>
<servlet-class>com.wrox.initParams</servlet-class>
<init-param>
<param-name>database</param-name>
<param-value>CustomerSupport</param-value>
</init-param>
<init-param>
<param-name>server</param-name>
<param-value>10.0.12.5</param-value>
</init-param>
</servlet>
<servlet-mapping>
<servlet-name>servletParameterServlet</servlet-name>
<url-pattern>/servletParameter</url-pattern>
</servlet-mapping>
</web-app>
这是映射到servletParameterServlet的页面。
package com.wrox;
import javax.servlet.http.HttpServlet;
import javax.servlet.ServletConfig;
import javax.servlet.ServletContext;
import javax.servlet.ServletException;
import javax.servlet.annotation.WebInitParam;
import javax.servlet.annotation.WebServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
import java.io.IOException;
import java.io.PrintWriter;
import java.util.Enumeration;
import java.util.List;
public class initParams extends HttpServlet{
@Override
protected void doGet(HttpServletRequest request, HttpServletResponse response)
throws ServletException, IOException
{
ServletContext c = this.getServletContext();
PrintWriter writer = response.getWriter();
Enumeration<String> temp = c.getInitParameterNames();
while(temp.hasMoreElements()) {
writer.append(temp.nextElement());
}
writer.append("database: ").append(c.getInitParameter("database"))
.append(", server: ").append(c.getInitParameter("server"));
}
@Override
public void init(ServletConfig config) throws ServletException
{
super.init(config);
System.out.println("Servlet " + this.getServletName() + " has started.");
}
@Override
public void destroy() {
System.out.println("Servlet " + this.getServletName() + " has stopped.");
}
}
答案 0 :(得分:0)
您初始化事物的方式,我认为您正在寻找的内容应该在
中提供this.getServletConfig().getInitParameter("foo");
一般来说,servlet上下文由JVM内的servlet共享,但是这个对象存在于特定servlet的servlet配置中,而<init-param>
则位于ServletConfig中。
有关详细信息,我强烈建议您阅读文档。