我正在自学JSP/Servlet。我面临着一个我能解决的问题。我正在创建一个请求servlet的简单表单。问题是,当我将web.xml中的url模式更改为我想要的url时,Tomcat会给我一个404错误。但是,当我将url模式更改为与servlet相同的名称时,它就可以工作了。我注意到的另一件事是,当我手动输入我想要的URL模式时,它可以工作。看来我没有被引导到正确的地方。我已经检查了很多次web.xml,我找不到任何错误。下面是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);
}
}
这里是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>
对你的所作所为有很多批评,但我将仅限于你的问题。
如果将应用程序部署到Tomcat 7/webapps目录下的WAR文件foo中。war,那么调用AddToEmailListServlet
并在浏览器中显示该HTML页面的URL将是:
http://host:8080/foo/AddToEmailListServlet
我假设您在表单中张贴这三个请求参数,因为您必须在发送电子邮件地址之前对@符号进行编码。
在"WebContent"目录下创建一个新文件夹,而不是像admin.xhtml那样给出特定的文件url。在这种情况下,假设文件夹名称是"安全的",将admin.xhtml放在该文件夹中,然后<url-pattern>/secured/*</url-pattern>
为我工作,我希望这对我有帮助