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JAVAservlet中的lang.NumberFormatException

这是一个简单的计算器,有两个运算:“+”和“-”。 不久前我运行它时,一切正常。但在某个时刻,我犯了错误,无法理解发生了什么

这是我的消息来源:

html

 <!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>Calculator</title>
</head>
<body>
    <h1>Calculator</h1>
    <form action="./" method="get">
        <label>First number:</label> <input type="text" name="num1" /> <br />
        <label>Second number: </label> <input type="text" name="num2" /> <br />
            <select name="operator">
                <option value="+">+</option>
                <option value="-">-</option>
            </select>
        <input type="submit" value="submit" />
    </form>
</body>
</html>

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>Calculator</display-name>
  <welcome-file-list>
    <welcome-file>index.html</welcome-file>
    <welcome-file>index.htm</welcome-file>
    <welcome-file>index.jsp</welcome-file>
    <welcome-file>default.html</welcome-file>
    <welcome-file>default.htm</welcome-file>
    <welcome-file>default.jsp</welcome-file>
  </welcome-file-list>
  <servlet>
    <servlet-name>Calculator</servlet-name>
    <servlet-class>ua.nure.nechyporuk.Calculator</servlet-class>
  </servlet>
  <servlet-mapping>
    <servlet-name>Calculator</servlet-name>
    <url-pattern>/</url-pattern>
  </servlet-mapping>
</web-app>

。爪哇

 package ua.nure.nechyporuk;

import java.io.*;
import javax.servlet.*;
import javax.servlet.http.*;



public class Calculator extends HttpServlet {

    private static final long serialVersionUID = 1L;

    public void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
        try {
            response.setContentType("text/html");
            int a1 = Integer.parseInt(request.getParameter("num1"));
            int a2 = Integer.parseInt(request.getParameter("num2"));
            String operator = request.getParameter("operator");
            if (operator.equals("+")&& operator != null) {
                 request.setAttribute("operation", "addition");
                 request.setAttribute("res", a1+a2);
                 request.getRequestDispatcher("index.jsp").forward(request, response);
            }
            if (operator.equals("-")&& operator != null) {
                request.setAttribute("operation", "subtract");
                 request.setAttribute("res", a1-a2);
                 request.getRequestDispatcher("index.jsp").forward(request, response);
            }

        } catch (Exception e) {
            e.printStackTrace();
        }
    }
}

jsp

 <%@ page contentType="text/html;charset=UTF-8" language="java" %>
<html>
<head>
    <title>Result of calculation:</title>
</head>
<body>
    <h1>Result of ${operation}</h1>
    <h1>${res}</h1>
   <a href= "http://localhost:8081/Calculator">back to calculator</a>;
</body>
</html>

我不明白为什么会出现这个错误。 就像html页面没有时间工作一样。 Tomkat I设置为端口8081

例外情况:

java.lang.NumberFormatException: For input string: "null"
    at java.lang.NumberFormatException.forInputString(NumberFormatException.java:65)
    at java.lang.Integer.parseInt(Integer.java:580)
    at java.lang.Integer.parseInt(Integer.java:615)
    at ua.nure.nechyporuk.Calculator.doGet(Calculator.java:17)
    at javax.servlet.http.HttpServlet.service(HttpServlet.java:624)
    at javax.servlet.http.HttpServlet.service(HttpServlet.java:731)
    at org.apache.catalina.core.ApplicationFilterChain.internalDoFilter(ApplicationFilterChain.java:303)
    at org.apache.catalina.core.ApplicationFilterChain.doFilter(ApplicationFilterChain.java:208)
    at org.apache.tomcat.websocket.server.WsFilter.doFilter(WsFilter.java:52)
    at org.apache.catalina.core.ApplicationFilterChain.internalDoFilter(ApplicationFilterChain.java:241)
    at org.apache.catalina.core.ApplicationFilterChain.doFilter(ApplicationFilterChain.java:208)
    at org.apache.catalina.core.StandardWrapperValve.invoke(StandardWrapperValve.java:218)
    at org.apache.catalina.core.StandardContextValve.invoke(StandardContextValve.java:122)
    at org.apache.catalina.authenticator.AuthenticatorBase.invoke(AuthenticatorBase.java:505)
    at org.apache.catalina.core.StandardHostValve.invoke(StandardHostValve.java:169)
    at org.apache.catalina.valves.ErrorReportValve.invoke(ErrorReportValve.java:103)
    at org.apache.catalina.valves.AccessLogValve.invoke(AccessLogValve.java:956)
    at org.apache.catalina.core.StandardEngineValve.invoke(StandardEngineValve.java:116)
    at org.apache.catalina.connector.CoyoteAdapter.service(CoyoteAdapter.java:442)
    at org.apache.coyote.http11.AbstractHttp11Processor.process(AbstractHttp11Processor.java:1082)
    at org.apache.coyote.AbstractProtocol$AbstractConnectionHandler.process(AbstractProtocol.java:623)
    at org.apache.tomcat.util.net.JIoEndpoint$SocketProcessor.run(JIoEndpoint.java:316)
    at java.util.concurrent.ThreadPoolExecutor.runWorker(ThreadPoolExecutor.java:1142)
    at java.util.concurrent.ThreadPoolExecutor$Worker.run(ThreadPoolExecutor.java:617)
    at org.apache.tomcat.util.threads.TaskThread$WrappingRunnable.run(TaskThread.java:61)
    at java.lang.Thread.run(Thread.java:745)

共 (1) 个答案

  1. # 1 楼答案

    而不是用这个

    Integer.parseInt(request.getParameter("num1"));
    

    试试这个。只需在request.getParameter("num1")之前添加"" +

    Integer.parseInt("" + request.getParameter("num1"));
    

    希望这能解决你的问题。并确保在字段中输入有效的整数

    更新

    检查null

        String str=request.getParameter("num1");
    
        int a=0;
    
        if(str!=null){
           a = Integer.parseInt("" + str);
        }