3

I am developing a netbeans web server with spring MVC in dispatcher-servlet.xml file the header that was generated automatically looks like this.

<?xml version='1.0' encoding='UTF-8' ?>
<beans xmlns="http://www.springframework.org/schema/beans"
       xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
       xmlns:p="http://www.springframework.org/schema/p"
       xmlns:aop="http://www.springframework.org/schema/aop"
       xmlns:tx="http://www.springframework.org/schema/tx"
       xsi:schemaLocation="http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans-4.0.xsd
       http://www.springframework.org/schema/aop http://www.springframework.org/schema/aop/spring-aop-4.0.xsd
       http://www.springframework.org/schema/tx http://www.springframework.org/schema/tx/spring-tx-4.0.xsd">

and in some part of the xml I have this

<bean id="viewResolver"
      class="org.springframework.web.servlet.view.InternalResourceViewResolver"
      p:prefix="/WEB-INF/jsp/"
      p:suffix=".jsp" />

the suffix p is used but when I access to the url http://www.springframework.org/schema/p it give me a 404

sorry but I'm new in java and I don't know too much

how can I fix this problem?

thanks in advance

Emiliano
  • 698
  • 9
  • 30
  • Why are you thinking that you need to "access the URL" at all? This isn't so much a Spring/Java question as it is an XML Schema question. (As a side note, manual XML-style configuration for Spring applications is mostly outdated. I suggest looking at the official Spring Getting Started Guides and using Spring Boot, which, for example, would register that bean for you with no configuration necessary.) – chrylis -cautiouslyoptimistic- May 14 '17 at 23:17
  • 1
    I need access to that url to avoid errors when I run the application the error say that need p – Emiliano May 14 '17 at 23:25
  • I have the same problem. Yesterday it worked, now P is not accessible anymore. – Brain May 18 '17 at 07:51

0 Answers0