Criando filtro no Web.xml

1 resposta
T

Saudações a todos

Galera, quando tento criar um filtro no meu web.xml, no fica aparecendo,
The content of element type "filter-mapping" must match "(filter-name,(url-pattern|servlet-name))".

Segue todo meu web.xml:

<?xml version="1.0" encoding="UTF-8"?>
<!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 id="WebApp_ID">
	<display-name>GTU</display-name>
	<servlet>
		<servlet-name>Faces Servlet</servlet-name>
		<servlet-class>javax.faces.webapp.FacesServlet</servlet-class>
		<load-on-startup>1</load-on-startup>
	</servlet>
	<servlet-mapping>
		<servlet-name>Faces Servlet</servlet-name>
		<url-pattern>/faces/*</url-pattern>
	</servlet-mapping>
	<servlet-mapping>
		<servlet-name>Faces Servlet</servlet-name>
		<url-pattern>*.jsf</url-pattern>
	</servlet-mapping>
	<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>
	<filter>
		<filter-name>filtroSeguranca</filter-name>
		<filter-class>gtu.util.Security</filter-class>
	</filter>
	<filter-mapping>
		<filter-name>filtroSeguranca</filter-name>
		<url-pattern>/*</url-pattern>
		<dispatcher>REQUEST</dispatcher>
		<dispatcher>FORWARD</dispatcher>
		<dispatcher>INCLUDE</dispatcher>
		<dispatcher>ERROR</dispatcher>
	</filter-mapping>
</web-app>

Saudações a todos, obrigado.

1 Resposta

Andre_Fonseca

Oi,

Tenta tirar esse <dispatcher> do <filter-mapping>

[]´s

Criado 11 de agosto de 2008
Ultima resposta 11 de ago. de 2008
Respostas 1
Participantes 2