2014-12-07 65 views
-1

以下是我的jsp代碼。使用Apache tomcat v7在Eclipse中的servlet中發生HTTP 404錯誤

<%@ page language="java" contentType="text/html; charset=ISO-8859-1" 
pageEncoding="ISO-8859-1"%> 
<%@page import="connection.*" %> 
<%@page import="java.sql.*" %> 

    <!DOCTYPE html> 
    <html> 
    <head> 
    <meta charset="ISO-8859-1"> 
    <title>VIEW REPORTS</title> 
    </head> 
    <body> 
    <div align="center"><b>View Reports</b></div> 
    <form action="GetRole" method="get"> 
    <p>Enter user id: 
    <input type="text" name="userid"><br> 
    <input type="submit" value="submit"><br> 
    </form> 
    Choose the type of report: 
    <script type="text/javascript" src="list.js"></script> 
    <select name="first1" id="firstselect" onchange="first()"> 

     <option>Select</option> 
     <option value="NEFT">NEFT REPORTS</option> 
     <option value="LOAN">LOAN REPORTS</option> 
     <option value="TRAN">TRANSACTION REPORTS</option> 
    </select> 

    <script type="text/javascript"></script> 


    </body> 
    </html> 

以下是我的servlet:

 package all_packages; 
     import connection.*; 
     import java.io.IOException; 
     import java.sql.*; 

     import javax.servlet.ServletException; 
     //import javax.servlet.annotation.WebServlet; 
     import javax.servlet.http.HttpServlet; 
     import javax.servlet.http.HttpServletRequest; 
     import javax.servlet.http.HttpServletResponse; 
     public class GetRole extends HttpServlet { 
     private static final long serialVersionUID = 1L; 
     protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException { 
     try { 
      Connection con=getConnection.getConnectionBuilder(); 
      String userid=request.getParameter("userid"); 
      System.out.println(" "+userid); 
      PreparedStatement pstmt=con.prepareStatement("select role_id from users where role_id=?"); 
      pstmt.setString(1,"userid"); 
      ResultSet rs=pstmt.executeQuery(); 
      while(rs.next()) 
      { 
       System.out.println(" "+rs.getString(1)); 
      } 
     } catch (SQLException e) { 
      System.out.println("Connection unsuccessful"); 
      e.printStackTrace(); 
     } 
     } 
     } 

以下是我web.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" xmlns:web="http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd" 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>Final_Bank</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>GetRole</servlet-name> 
     <servlet-class>GetRole</servlet-class> 
    </servlet> 
    <servlet-mapping> 
     <servlet-name>GetRole</servlet-name> 
     <url-pattern>/GetRole</url-pattern> 
    </servlet-mapping> 
    </web-app> 

請讓我知道爲什麼它拋出404錯誤。我在網上搜索了很長時間纔得到正確的解決方案,但沒有解決方案爲我工作。我在Eclipse中使用了Apache Tomcat服務器。

+0

何時發生錯誤?哪個網址?提交表單後? – andih 2014-12-07 08:29:41

+0

@andih,提交表單後回覆。 Aftrer提交,URL即將到來:'http:// localhost:8080/Final_Bank/GetRole?userid = wcw' – Mistu4u 2014-12-07 09:07:06

+0

您可以更新您的文章中的'web.xml'來表示您當前的本地版本。 – andih 2014-12-07 09:14:24

回答

0

該servlet是在包all_packages所以在web.xml文件映射應該是:

<servlet> 
    <servlet-name>GetRole</servlet-name> 
    <servlet-class>all_packages.GetRole</servlet-class> 
</servlet> 

現在來到404問題,如果JSP名稱是說sample.jsp和應用程序上下文名稱爲myapp則你需要訪問的網址是:

http://localhost:8080/myapp/sample.jsp 

假設你的Tomcat是在端口8080運行。

+0

如果這是我期望在WebContainer/WebApplication/Tomcat啓動期間出現'ClassNotFoundException'的原因。 – andih 2014-12-07 08:37:40

+0

此修改後仍然出現404錯誤。 – Mistu4u 2014-12-07 09:06:19

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