手动实现 Tomcat 底层机制+ 自己设Servlet问题分析

完成小案例

运行效果

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此项目用maven至于怎么配置在下一篇文章

创建cal.html

<!DOCTYPE html>
<html lang="en"><head><meta charset="UTF-8"><title>计算器</title>
</head><body><h1>计算器</h1><form action="/calServlet" method="get">num1:<input type="text" name="num1"><br />num2:<input type="text" name="num2"><br /><input type="submit" value="提交"></form>
</body></html>

CalServlet.java

# 实现步骤

1.接收提交的数据进行计算
2.复制当前行 ctrl+alt+下光标
3.把strNum1 和 strNum2 转成 int
4.写入数据发送前端
5.关闭

public class CalServlet extends HttpServlet {protected void doPost(HttpServletRequest request,HttpServletResponse response) throws ServletException, IOException {//接收提交的数据进行计算//复制当前行 ctrl+alt+下光标String strNum1 = request.getParameter("num1");String strNum2 = request.getParameter("num2");//把strNum1 和 strNum2 转成 intint num1 = WebUtils.parseInt(strNum1, 0);int num2 = WebUtils.parseInt(strNum2, 0);int result = num1 + num2;response.setContentType("text/html;charset=utf-8");PrintWriter writer = response.getWriter();//写入数据发送前端writer.print("<h1>" + num1 + " + " + num2 + " = " + result + "<h1>");writer.flush();//注意一定要关闭不然有可能提交失败writer.close();}protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {doPost(request, response);}
}

web.xml

注意
servlet-mapping的servlet-nameCalServlet对应的是servlet下的servlet-name的CalServlet名字注意这名字是需要一样的

下面servlet-class指的是对应这那个类来处理/calServlet是对应这前端html的表单注意需要一样的否则无法拦截
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servlet-mapping的url-pattern

<!DOCTYPE web-app PUBLIC"-//Sun Microsystems, Inc.//DTD Web Application 2.3//EN""http://java.sun.com/dtd/web-app_2_3.dtd" >
<web-app><display-name>Archetype Created Web Application</display-name><servlet><servlet-name>CalServlet</servlet-name><servlet-class>com.servlet.CalServlet</servlet-class></servlet><servlet-mapping><servlet-name>CalServlet</servlet-name><url-pattern>/calServlet</url-pattern></servlet-mapping>
</web-app>

WebUtils

将一个字符串数字，转成 int, 如果转换失败，就返回传入 defaultVal

public class WebUtils {/*** 将一个字符串数字，转成 int, 如果转换失败，就返回传入 defaultVal* @param strNum* @param defaultVal* @return*/public static int parseInt(String strNum, int defaultVal) {try {return Integer.parseInt(strNum);} catch (NumberFormatException e) {System.out.println(strNum + " 格式不对，转换失败");}return defaultVal;}}