tomcat - how to create War file for a RESTful web service developed in eclipse IDE -
i ve created sample rest web service writes data in xml file. have hard coded path xml file written. want know how declare local path of file in web.xml file servlet parameter , how path there , use in codebe . need create war file service needs deployed in tomcat. war file should take parameter web.xml file. used eclipse ide develop web service. can tell me how above things ?
here have attached servlet code present inside web.xml file.
<servlet> <servlet-name>jersey rest service</servlet-name> <servlet-class> com.sun.jersey.spi.container.servlet.servletcontainer </servlet-class> <init-param> <param-name>com.sun.jersey.config.property.packages</param-name> <param-value>com.sample.service</param-value> </init-param> <init-param> <param-name>filepath</param-name> <param-value>filepath value</param-value> </init-param> <load-on-startup>1</load-on-startup> </servlet> <servlet-mapping> <servlet-name>jersey rest service</servlet-name> <url-pattern>/api/*</url-pattern> </servlet-mapping> com.sample.service package have rest web service class.
assuming created dynamic web project in eclipse, right-click on
project name, > export > war file
and fill in details asks for.
in web.xml, can define filepath below
<servlet> <servlet-name>myservletname</servlet-name> <servlet-class>com.mycompany.myservlet</servlet-class> <init-param> <param-name>filepath</param-name> <param-value>d:\hard-coded-path.xml</param-value> </init-param> </servlet> *updated correct answer per comments *
you're getting nullpointerexception on getservletcontext().getinitparameter("filepath") because context not injected web service method.
and in web service, use code path , write using @context annotation
@get @produces("text/plain") public string dostuff(@context servletconfig sc) { string xmlpath = "output filepath is: " + sc.getinitparameter("filepath"); return xmlpath; }
Comments
Post a Comment