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HTTP状态500-NoSuchElementException

[英]HTTP Status 500 - NoSuchElementException

我一直收到此错误,但还没有找到解决它的方法。 似乎与扫描仪有关,但我不确定此处要更改什么。

我的LoginServlet:

package web;

import java.io.BufferedWriter;
import java.io.FileReader;
import java.io.FileWriter;
import java.io.IOException;
import java.io.PrintWriter;
import java.util.Scanner;

import javax.servlet.RequestDispatcher;
import javax.servlet.ServletException;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;

public class LoginServlet extends HttpServlet {
    protected void doPost(HttpServletRequest req, HttpServletResponse resp)
            throws ServletException, IOException {
        // ______________________________________________________________
        boolean loginSuccess = false;
        String naam = "";
        String pass = "";
        RequestDispatcher rd = null;
        String operator = req.getParameter("operator");

        if (operator.equals("Register")) {
            req.getRequestDispatcher("Register.jsp").forward(req, resp);
        }
        // _________________________*******_________________________________

        // ___________________scanner declaratie________
        FileReader fr = new FileReader("users.txt");
        Scanner sc = new Scanner(fr);
        // _________________________*******____________

        // ___________________!! Lezen van file !!________________________
        if (operator.equals("Login")) {
            while (sc.hasNextLine()) {
                String regel = sc.nextLine();
                sc.useDelimiter("\\s:\\s");
                naam = sc.next();
                pass = sc.next();

                System.out.println(naam);
                System.out.println(pass);
                // ___________________Toevoegen aan logged users_____
                if (req.getParameter("username").equals(naam)
                        && (req.getParameter("password").equals(pass))) {
                    PrintWriter out = new PrintWriter(new BufferedWriter(
                            new FileWriter("loggedusers.txt", true)));

                    out.println(naam);
                    out.close();
                    loginSuccess = true;
                    break;
                } else if (req.getParameter("username").equals(naam)
                        && (!req.getParameter("password").equals(pass))) {
                    req.setAttribute("error", "Password incorrect");
                    loginSuccess = false;
                    break;
                } else if (!sc.hasNextLine()) {
                    req.setAttribute("error", "Username doesn't exist");
                    loginSuccess = false;
                    break;
                }
            }
            sc.close();

            if (loginSuccess = true) {
                req.getRequestDispatcher("Welcome.jsp").forward(req, resp);
            } else if (loginSuccess = false) {
                req.getRequestDispatcher("Login.jsp").forward(req, resp);
            }


        }
    }
}

我的Login.jsp:

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN"
"http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="content-type" content="text/html; charset=iso-8859-1" />
<title>Login page</title>
</head>
<body>
    <form action="LoginServlet.do" method="post">
        <table>
            <tr>
                <td align="right">Username:</td>
                <td align="left"><input type="text" name="username" /></td>
            </tr>
            <tr>
                <td align="right">Password:</td>
                <td align="left"><input type="text" name="password" /></td>
            </tr>
            <tr>
                <td align="left"><input type="submit" name="operator" value="Register" /></td>
                <td align="right"><input type="submit" name="operator" value="Login" /></td>
            </tr>
        </table>

    </form>
    <h1><%=request.getAttribute("error") %></h1>
</body>
</html>

web.xml应该是正确的,因为它以前可以工作,但是在从servlet编写其余代码之后,它会卡住。

它在Eclipse中给我的错误:

jun 03, 2015 4:44:44 PM org.apache.catalina.core.StandardWrapperValve invoke
SEVERE: Servlet.service() for servlet [Dynamic Example Servlet] in context with path [/DynamicExample] threw exception
java.util.NoSuchElementException
    at java.util.Scanner.throwFor(Unknown Source)
    at java.util.Scanner.next(Unknown Source)
    at web.LoginServlet.doPost(LoginServlet.java:41)
    at javax.servlet.http.HttpServlet.service(HttpServlet.java:648)
    at javax.servlet.http.HttpServlet.service(HttpServlet.java:729)
    at org.apache.catalina.core.ApplicationFilterChain.internalDoFilter(ApplicationFilterChain.java:291)
    at org.apache.catalina.core.ApplicationFilterChain.doFilter(ApplicationFilterChain.java:206)
    at org.apache.tomcat.websocket.server.WsFilter.doFilter(WsFilter.java:52)
    at org.apache.catalina.core.ApplicationFilterChain.internalDoFilter(ApplicationFilterChain.java:239)
    at org.apache.catalina.core.ApplicationFilterChain.doFilter(ApplicationFilterChain.java:206)
    at org.apache.catalina.core.StandardWrapperValve.invoke(StandardWrapperValve.java:219)
    at org.apache.catalina.core.StandardContextValve.invoke(StandardContextValve.java:106)
    at org.apache.catalina.authenticator.AuthenticatorBase.invoke(AuthenticatorBase.java:502)
    at org.apache.catalina.core.StandardHostValve.invoke(StandardHostValve.java:142)
    at org.apache.catalina.valves.ErrorReportValve.invoke(ErrorReportValve.java:79)
    at org.apache.catalina.valves.AbstractAccessLogValve.invoke(AbstractAccessLogValve.java:617)
    at org.apache.catalina.core.StandardEngineValve.invoke(StandardEngineValve.java:88)
    at org.apache.catalina.connector.CoyoteAdapter.service(CoyoteAdapter.java:518)
    at org.apache.coyote.http11.AbstractHttp11Processor.process(AbstractHttp11Processor.java:1091)
    at org.apache.coyote.AbstractProtocol$AbstractConnectionHandler.process(AbstractProtocol.java:668)
    at org.apache.tomcat.util.net.NioEndpoint$SocketProcessor.doRun(NioEndpoint.java:1521)
    at org.apache.tomcat.util.net.NioEndpoint$SocketProcessor.run(NioEndpoint.java:1478)
    at java.util.concurrent.ThreadPoolExecutor.runWorker(Unknown Source)
    at java.util.concurrent.ThreadPoolExecutor$Worker.run(Unknown Source)
    at org.apache.tomcat.util.threads.TaskThread$WrappingRunnable.run(TaskThread.java:61)
    at java.lang.Thread.run(Unknown Source)

以及运行时在浏览器中看到的屏幕截图: http://puu.sh/iaWY9/5b0da53a66.png

我希望收到你们的来信! 预先加油,谢谢!

三思而后行。 或更准确地说, 请勿将Scanner提前两次。

这些是罪魁祸首:

String regel = sc.nextLine();
naam = sc.next();
pass = sc.next();

hasNextLine()对调用nextLine() ,但对随后的next()调用无效。

可能想做的是改为从regel读取naampaas值。

String regel = sc.nextLine();
String[] regelBroken = regel.split();
naam = regelBroken[0];
paas = regelBroken[1];

这样,您只需将扫描仪推进一次,就不会遇到无法进行扫描的情况。

在“ users.txt ”的末尾是否有空行?

在最后一次迭代中,这可能会导致naam = sc.next(); 抛出异常。

我想通了,我尝试了一些不同:

 FileReader fr = new FileReader("users.txt"); BufferedReader br = new BufferedReader(fr); while (true) { String regel = br.readLine(); if (regel == null) { break; } sc = new Scanner(regel); sc.useDelimiter("\\\\s:\\\\s"); naam = sc.next(); pass = sc.next(); 

有用!

感谢数组的解决方案,我保存了这段代码,我想以后可以尝试这种方法,谢谢大家!

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