Hi,
Can you show me the camel-config.xml?
A simple demo help me debug the issue.
Willem
janylj wrote:
willem.jiang wrote:
If you create the servlet component, you need to specify the servlet
which the component will attach to.
<bean id="servlet"
class="org.apache.camel.component.servlet.ServletComponent">
<property name="camelContext" ref="camel" />
<property name="servletName" value="CamelServlet" />
</bean>
doesn't work, because Tomcat starts error said,
Caused by: org.springframework.beans.InvalidPropertyException: Invalid
property 'servletName' of bean class
[org.apache.camel.component.servlet.ServletComponent]: No property
'servletName' found
willem.jiang wrote:
You don't need to specify the servlet component in spring, if there is
only one camel servlet created in the web context.
doesn't work, because Tomcat starts error said,
Caused by: java.lang.IllegalArgumentException: Can't find the deployied
servlet, please set the ServletComponent with it or delopy a
CamelHttpTransportServlet int the web container
What I am missing? Please note that I am using Spring ContextLoaderListener.
Thanks a lot.
my web.xml,
<!-- location of spring xml files -->
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>classpath:camel-config.xml</param-value>
</context-param>
<!-- the listener that kick-starts Spring -->
<listener>
<listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>
<servlet>
<servlet-name>CamelServlet</servlet-name>
<servlet-class>
org.apache.camel.component.servlet.CamelHttpTransportServlet
</servlet-class>
<init-param>
<param-name>matchOnUriPrefix</param-name>
<param-value>true</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>CamelServlet</servlet-name>
<url-pattern>/services/*</url-pattern>
</servlet-mapping>