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Gabriel vargas
How can I call a Servlet that is declare on web.xml with spring framework
0 Answer(s)      2 years and 4 months ago
Posted in : Spring

here is my problem, I have spring framework and Crystal Report

the problem is when Report.rpt call the servlet CrystalReportViewerServlet the servlet is declare on web.xml, but the spring dispacher don't find it so how I can make spring dispacher to call a servlet that is declare on my Web.xml

here is my web.xml

<servlet>
    <servlet-name>CrystalReportViewerServlet</servlet-name>
    <servlet-class>com.crystaldecisions.report.web.viewer.CrystalReportViewerServlet</servlet-class>
  </servlet>
  <servlet-mapping>
    <servlet-name>CrystalReportViewerServlet</servlet-name>
    <url-pattern>/CrystalReportViewerHandler</url-pattern>
  </servlet-mapping>


  <context-param>
    <param-name>crystal_image_uri</param-name>
    <param-value>/gaa.vargas.furniture/crystalreportviewers</param-value>
  </context-param>
  <context-param>
    <param-name>crystal_image_use_relative</param-name>
    <param-value>webapp</param-value>
  </context-param>

  <context-param>
    <param-name>contextConfigLocation</param-name>
    <param-value>
                /WEB-INF/applicationContext.xml
        </param-value>
  </context-param>

<listener>
    <listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
  </listener>
  <servlet>
    <servlet-name>dispatcher</servlet-name>
    <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
    <load-on-startup>1</load-on-startup>
  </servlet>
  <servlet-mapping>
    <servlet-name>dispatcher</servlet-name>
    <url-pattern>*.htm</url-pattern>
  </servlet-mapping>
  <session-config>
    <session-timeout>30</session-timeout>
  </session-config>
  <welcome-file-list>
    <welcome-file>redirect.jsp</welcome-file>
  </welcome-file-list>
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