我已經寫了一個servlet程序,用於登錄頁面..am使用tomcat服務器...後,我在服務器上運行得到上述錯誤... 下面是我的servlet代碼。HTTP狀態404 - /請求的資源不可用
包演示;
import java.io.IOException;
import java.io.PrintWriter;
import java.sql.Connection;
import java.sql.DriverManager;
import java.sql.ResultSet;
import java.sql.SQLException;
import java.sql.Statement;
import javax.servlet.ServletException;
import javax.servlet.annotation.WebServlet;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
/**
* Servlet implementation class LoginServlet
*/
public class LoginServlet extends HttpServlet {
static private String dbUrl="jdbc:mysql://localhost:3306/employee";
static private String dbUn="root";
static private String dbPwd="root";
static private Connection ConObj;
static private Statement StmtObj;
static private ResultSet RsObj;
public void service(HttpServletRequest request,HttpServletResponse response)throws IOException
{
try {
Class.forName("com.mysql.jdbc.driver");
ConObj=DriverManager.getConnection(dbUrl, dbUn, dbPwd);
StmtObj=ConObj.createStatement();
response.setContentType("text/html");
PrintWriter out=response.getWriter();
out.write("<html><body>");
out.write("<h2>");
String ActLogName=request.getParameter("Logname");
String ActPwd=request.getParameter("Pwd");
String SqlQuery="select * from users where username='"+ActLogName+"' and password='"+ActPwd+"'";
RsObj=StmtObj.executeQuery(SqlQuery);
if(RsObj.next()==true)
{
String ExpLogName=RsObj.getString("username");
String ExpPwd=RsObj.getString("password");
if(ActLogName.equals(ExpLogName)&& ActPwd.equals(ExpPwd))
{
out.write("Login Success");
}
}
else
{
out.write("Login Failed");
}
out.write("</h2>");
out.write("</body></html>");
}
catch (ClassNotFoundException|SQLException exp) {
exp.printStackTrace();
}
finally{
try {
RsObj.close();
StmtObj.close();
ConObj.close();
} catch (SQLException e) {
e.printStackTrace();
}
}
}
}
下面是我的html代碼
<!DOCTYPE html>
<html>
<head>
<title>login page</title>
</head>
<body>
<form action="http://localhost:8080/FlipKart/loginpage">
LoginName :<input type="text" name="Logname"><br>
Password :<input type="password" name="Pwd"><br>
<input type="submit" value="Login">
<input type="button" value="cancel">
</form>
</body>
</html>
下面是我的web.xml代碼
<web-app>
<servlet>
<servlet-name>loginserv</servlet-name>
<servlet-class>LoginServlet</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>loginserv</servlet-name>
<url-pattern>/loginpage</url-pattern>]
</servlet-mapping>
</web-app>
有人可以幫我去解決這個問題.. 在此先感謝.. :)
你想要打什麼URL? –
您需要確保部署在Tomcat上的應用程序被命名爲FlipKart。 –
另一件事是,如果您訪問的頁面和操作中的servlet託管在同一位置,則您應該在表單標記的操作屬性中使用相對URL。 –