我正在自学JSP/Servlet。我正面临一个我能够解决的问题。 我正在创建一个将请求 servlet 的简单表单。问题是当我将 web.xml 中的 url-pattern 更改为我想要的 url 时，Tomcat 给我一个错误 404。但是，当我将 url-pattern 更改为与它工作的 servlet 相同的名称时。我注意到的另一件事是，当我在 URL 上手动键入我想要的 url-pattern 时，它会起作用。 看来我没有被重定向到正确的地方。我已经多次检查 web.xml，但没有发现任何错误。这是 servlet 代码:
package email;
import java.io.*;
import javax.servlet.*;
import javax.servlet.http.*;
import business.User;
import data.UserIO;
/**
* @author Joel Murach
*/
public class AddToEmailListServlet extends HttpServlet
{
int globalCount;
public void init() throws ServletException{
globalCount = 0;
}
protected void doPost(
HttpServletRequest request,
HttpServletResponse response)
throws ServletException, IOException
{
//Global variable
globalCount++;
// get parameters from the request
String firstName = request.getParameter("firstName");
String lastName = request.getParameter("lastName");
String emailAddress = request.getParameter("emailAddress");
// get a relative file name
ServletContext sc = getServletContext();
String path = sc.getRealPath("/WEB-INF/EmailList.txt");
// use regular Java objects to write the data to a file
User user = new User(firstName, lastName, emailAddress);
UserIO.add(user, path);
// send response to browser
response.setContentType("text/html;charset=UTF-8");
PrintWriter out = response.getWriter();
out.println(
"<!doctype html public \"-//W3C//DTD HTML 4.0 Transitional//EN\">\n"
+ "<html>\n"
+ "<head>\n"
+ " <title>Murach's Java Servlets and JSP</title>\n"
+ "</head>\n"
+ "<body>\n"
+ "<h1>Thanks for joining our email list</h1>\n"
+ "<p>Here is the information that you entered:</p>\n"
+ " <table cellspacing=\"5\" cellpadding=\"5\" border=\"1\">\n"
+ " <tr><td align=\"right\">First name:</td>\n"
+ " <td>" + firstName + "</td>\n"
+ " </tr>\n"
+ " <tr><td align=\"right\">Last name:</td>\n"
+ " <td>" + lastName + "</td>\n"
+ " </tr>\n"
+ " <tr><td align=\"right\">Email address:</td>\n"
+ " <td>" + emailAddress + "</td>\n"
+ " </tr>\n"
+ " </table>\n"
+ "<p>To enter another email address, click on the Back <br>\n"
+ "button in your browser or the Return button shown <br>\n"
+ "below.</p>\n"
+ "<form action=\"join_email_list.html\" "
+ " method=\"post\">\n"
+ " <input type=\"submit\" value=\"Return\">\n"
+ "</form>\n"
+ "<p>This page has been accessed "
+ globalCount + " times.</p>"
+ "</body>\n"
+ "</html>\n");
System.out.println(globalCount);
log("Global variable" +globalCount);
out.close();
}
protected void doGet(
HttpServletRequest request,
HttpServletResponse response)
throws ServletException, IOException
{
doPost(request, response);
}
}
这是 web.xml 文件:
<?xml version="1.0" encoding="UTF-8"?>
<web-app version="2.5" xmlns="http://java.sun.com/xml/ns/javaee" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd">
<!-- the definitions for the servlets -->
<!-- the mapping for the servlets -->
<servlet>
<servlet-name>DisplayMusicChoicesServlet</servlet-name>
<servlet-class>email.DisplayMusicChoicesServlet</servlet-class>
</servlet>
<servlet>
<servlet-name>AddToEmailListServlet</servlet-name>
<servlet-class>email.AddToEmailListServlet</servlet-class>
</servlet>
<!-- other configuration settings for the application -->
<servlet-mapping>
<servlet-name>DisplayMusicChoicesServlet</servlet-name>
<url-pattern>/displayMusicChoices</url-pattern>
</servlet-mapping>
<servlet-mapping>
<servlet-name>AddToEmailListServlet</servlet-name>
<url-pattern>/addToEmailList</url-pattern>
</servlet-mapping>
<session-config>
<session-timeout>30</session-timeout>
</session-config>
<welcome-file-list>
<welcome-file>join_email_list.html</welcome-file>
</welcome-file-list>
</web-app>
最佳答案
很多人对你所做的事情提出批评，但我会限制自己回答你的问题。
如果您将应用程序部署到名为 foo.war 的 WAR 文件中的 Tomcat 7/webapps 目录，那么调用您的 AddToEmailListServlet 并在浏览器中显示该 HTML 页面的 URL 将是:
http://host:8080/foo/AddToEmailListServlet
我假设您在表单中发布这三个请求参数，因为您必须在发送之前对电子邮件地址中的 at 符号进行编码。
关于java - url 模式 Servlet Tomcat 7.0，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/8440093/