Java Rest API method throws a 404 Not Found after publishing

Question

I am hosting my Backend on Google App Engine and the Rest API works well on localhost. The method url is as follows:

http://127.0.0.1:8888/rest/plans/getplans/242353

This works on local host and returns me a JSON response properly. It's a GET method and 242353 is the parameter I pass to the method.

But when I publish it to my Google Cloud, after I attempt to call it on my browser with the following URL:

http://2-dot-MY_APP_DOMAIN.appspot.com/rest/plans/getplans/242353

I get this error on my browser:

Error: Not Found

The requested URL /rest/plans/getplans/242353 was not found on this server.

web.xml

<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
         xsi:schemaLocation="http://java.sun.com/xml/ns/javaee 
              http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd"
         version="2.5"
         xmlns="http://java.sun.com/xml/ns/javaee">


<servlet>
    <servlet-name>Jersey Web Application</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.dinukapj.socialapp.api</param-value>
     </init-param>
     <load-on-startup>1</load-on-startup>
  </servlet>
  <servlet-mapping>
    <servlet-name>Jersey Web Application</servlet-name>
    <url-pattern>/resources/*</url-pattern>
  </servlet-mapping>
   <servlet-mapping>
    <servlet-name>Jersey Web Application</servlet-name>
    <url-pattern>/rest/*</url-pattern>
  </servlet-mapping>
</web-app>

Why does this happen? Do I need to add any other information to my web.xml?

Answer 1

尝试按web.xml中的定义访问/ resources / plans / getplans / 242353