1

On my web.xml I have the following code

<servlet>
    <servlet-name>controller</servlet-name>
    <servlet-class>ControllerServlet</servlet-class>
    <load-on-startup>1</load-on-startup>
</servlet>


<servlet-mapping>
    <servlet-name>controller</servlet-name>
    <url-pattern>*.do</url-pattern>
</servlet-mapping>
<servlet-mapping>
    <servlet-name>controller</servlet-name>
    <url-pattern>/url/*</url-pattern>
</servlet-mapping>

To enter into the servlet the url has to be like : htp://ee:8080/ER/index.do or htp://ee:8080/ER/url/888. I want to get enter by a route like htp://ee:8080/94353.

I've tried a lot of possibilities but I can't find the correct one.

Thank you!

BalusC
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Dani Tome
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  • try with `/*.do` instead of `*.do`. Also you should provide the full name of the servlet (i.e. with the package). – bachr Jan 02 '15 at 17:57
  • @Arbi: Where did you learn that this pattern is supported? And, how exactly does it match OP's requirement? – BalusC Jan 02 '15 at 18:16
  • @BalusC `*.do` will match things like htp://ee:8080/ER/index.do and htp://ee:8080/ER/url/index.do (anything that ends with do). – bachr Jan 02 '15 at 18:27

1 Answers1

-2

I think you can't omit the application name from URL because in the URL you need to define the IP address of the server and the port number, in server there are a lot of applications are deployed in it. you need to tell server which application you want to access

abdotalaat
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