Acho que ou vc altera aqui :
<servlet-mapping>
<servlet-name>LoginServlet</servlet-name>
<url-pattern>/servlet/LoginServlet</url-pattern>
</servlet-mapping>
Ou aqui :
http://localhost:8080/myApp/LoginServlet
Rafael Tabosa
--- Igo Coelho <[EMAIL PROTECTED]> escreveu:
> Galera da lista estou tendo problemas para executar
> um servlet.
> Estou usando o Tomcat 5.0.28 e J2SDK 1.4.2_04. vou
> tentar ser bem
> claro para que voc�s possam me ajudar.
> Primeiro criei o diret�rio myApp/WEB-INF/classes
> dentro de
> CATALINA_HOME/webapps.
>
> Para criar meus servlets usei o JCreator. Para poder
> compilar fui a
> op��o de menu Configure/Options na op��o JDK
> Profiles editei jssdk1.4.2_04
> e adicionei os seguintes arquivos:
>
> D:\Tomcat 5.0\common\lib\servlet-api.jar
> D:\Tomcat 5.0\common\lib\jsp-api.jar
> D:\Tomcat 5.0\common\lib\tools.jar
>
> O c�digo do meu servlet � o seguinte:
> import javax.servlet.*;
> import javax.servlet.http.*;
> import java.io.*;
> public class LoginServlet extends HttpServlet{
>
> private void sendLoginForm(HttpServletResponse
> response, boolean withErrorMessage)
> throws ServletException,IOException{
>
> response.setContentType("text/html");
> PrintWriter out = response.getWriter();
>
> out.println("<html>");
> out.println("<head>");
> out.println("<title>Login</title>");
> out.println("</head>");
> out.println("<body>");
>
> if(withErrorMessage)
> out.println("Login failed. Please try
> again.<br>");
>
> out.println("<br>");
> out.println("<br>Please enter your user name and
> password.");
> out.println("<br><form action=post>");
> out.println("<br>User Name:<input type=text
> name=userName>");
> out.println("<br>Password:<input type=text
> name=password>");
> out.println("<br><input type=submit
> value=Submit>");
> out.println("</form>");
> out.println("</body>");
> out.println("</html>");
> }
>
> public void doGet(HttpServletRequest request,
> HttpServletResponse response)
> throws ServletException,IOException{
>
> sendLoginForm(response,false);
> }
>
> public void doPost(HttpServletRequest request,
> HttpServletResponse response)
> throws ServletException,IOException{
>
> String userName =
> request.getParameter("userName");
> String password =
> request.getParameter("password");
>
> if(userName != null && password != null &&
> userName.equals("Higor") &&
> password.equals("coelho")){
> response.sendRedirect("http://localhost:8080";);
> }else{
> sendLoginForm(response,true);
> }
> }
> }
>
> O c�digo de meu arquivo web.xml do diret�rio WEB-INF
> �:
>
> <?xml version="1.0" encoding="ISO-8859-1"?>
> <!DOCTYPE web-app
> PUBLIC "-//Sun Microsystems, Inc.//DTD Web
> Application 2.3//EN"
> "http://java.sun.com/dtd/web-app_2_3.dtd";>
> <web-app>
> <servlet>
> <servlet-name>LoginServlet</servlet-name>
> <servlet-class>LoginServlet</servlet-class>
> </servlet>
>
> <servlet-mapping>
> <servlet-name>LoginServlet</servlet-name>
> <url-pattern>/LoginServlet</url-pattern>
> </servlet-mapping>
> </web-app>
>
> No meu arquivo server.xml da pasta conf no diret�rio
> CATALINA_HOME
> eu inclui a seguinte linha antes do fechamento de
> tag </host>
>
> <Context path="/myApp" docBase="myApp" debug="0"
> privileged="true">
> </Context>
>
> Quando tento executar
> http://localhost:8080/myApp/servlet/LoginServlet
> o seguinte erro � mostrado:
>
> HTTP Status 404 - /myApp/servlet/LoginServlet
> type Status report
> message /myApp/servlet/LoginServlet
> description The requested resource
> (/myApp/servlet/LoginServlet) is not available.
> Apache Tomcat/5.0.28
>
>
> Alguem pode me dizer o que h� de errado?
>
>
>
>
> ---------------------------------
> Yahoo! Acesso Gr�tis - Internet r�pida e gr�tis.
> Instale o discador do Yahoo! agora.
_______________________________________________________
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