Question

因此，我是servlet和jsp的新手，并且紧跟https://stackoverflow.com/tags/servlets/info中的Hello World后处理示例。当我尝试在Eclipse中使用Tomcat v9.0运行它时，我得到了

在进行了一些其他研究并四处寻找之后，我还没有找到可行的解决方案或对正在发生的事情的解释。我基本上已经从示例中复制了代码，所以对于为什么它不起作用我感到很困惑。我也还没有足够的能力来弄清楚到底哪里出了问题，因此任何帮助都将是非常有用的。到目前为止，我唯一的预感是我可能弄乱了目录或其他内容。这是一张图片：

我唯一能找到的差异是我的HelloServlet.class所在的位置

 apache-tomcat-9.0.19/webapps/hello/build/classes/com/example/controller/HelloServlet.class

代替

/WEB-INF/classes/com/example/controller/HelloServlet.class

如示例中所述。我以为这是因为Eclipse默认情况下已经编译了现在的类文件，但是可以肯定的是，我将类文件夹复制到WEB-INF中，因此它与示例匹配，但是仍然无法正常工作。这就是我遇到的问题。如果有人能指出我的错误甚至根本没有帮助，那将非常感激。我在下面包括了hello.jsp，web.xml和HelloServlet.java文件，以防万一它们之间有任何问题。

hello.jsp

<%@ taglib uri="http://java.sun.com/jsp/jstl/functions" prefix="fn" %>
<!DOCTYPE html>
<html lang="en">
    <head>
        <title>Servlet Hello World</title>
        <style>.error { color: red; } .success { color: green; }</style>
    </head>
    <body>
        <form action="hello" method="post">
            <h1>Hello</h1>
            <p>
                <label for="name">What's your name?</label>
                <input id="name" name="name" value="${fn:escapeXml(param.name)}">
                <span class="error">${messages.name}</span>
            </p>
            <p>
                <label for="age">What's your age?</label>
                <input id="age" name="age" value="${fn:escapeXml(param.age)}">
                <span class="error">${messages.age}</span>
            </p>
            <p>
                <input type="submit">
                <span class="success">${messages.success}</span>
            </p>
        </form>
    </body>
</html>

web.xml

<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_4_0.xsd"
  version="4.0"
  metadata-complete="true">
</web-app>

HelloServlet.java

package com.example.controller;

import java.io.IOException;
import java.util.HashMap;
import java.util.Map;

import javax.servlet.ServletException;
import javax.servlet.annotation.WebServlet;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;

@SuppressWarnings("serial")
@WebServlet("/hello")
public class HelloServlet extends HttpServlet {

    @Override
    protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
        // Preprocess request: we actually don't need to do any business stuff, so just display JSP.
        request.getRequestDispatcher("/WEB-INF/hello.jsp").forward(request, response);
    }

    @Override
    protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
        // Postprocess request: gather and validate submitted data and display the result in the same JSP.

        // Prepare messages.
        Map<String, String> messages = new HashMap<String, String>();
        request.setAttribute("messages", messages);

        // Get and validate name.
        String name = request.getParameter("name");
        if (name == null || name.trim().isEmpty()) {
            messages.put("name", "Please enter name");
        } else if (!name.matches("\\p{Alnum}+")) {
            messages.put("name", "Please enter alphanumeric characters only");
        }

        // Get and validate age.
        String age = request.getParameter("age");
        if (age == null || age.trim().isEmpty()) {
            messages.put("age", "Please enter age");
        } else if (!age.matches("\\d+")) {
            messages.put("age", "Please enter digits only");
        }

        // No validation errors? Do the business job!
        if (messages.isEmpty()) {
            messages.put("success", String.format("Hello, your name is %s and your age is %s!", name, age));
        }

        request.getRequestDispatcher("/WEB-INF/hello.jsp").forward(request, response);
    }

}

Answer 1

您需要允许交叉原点

private void setAccessControlHeaders(HttpServletResponse resp) { resp.setHeader("Access-Control-Allow-Origin", "http://localhost:9000"); resp.setHeader("Access-Control-Allow-Methods", "GET"); }

如何修复错误404：原始服务器未找到目标资源的当前表示形式

1 个答案: