CODEWITHSUNDEEP
javaurlservletsservlet-filters
JAVAGeek
JAVAGeek

Reputation: 2794

Applying Filter on dynamic URL's

I have to filter urls like this :

http://www.mysite.com/708e93

where 708e93 is a unique id, based on this id i have to create dynamic pages in jsp.

Now the problem is - what do i put in url-pattern under filter-mapping

<filter-mapping>
<filter-name>foo.bar.MyFilter</filter-name>
<url-pattern> ?? </url-pattern>
</filter-mapping>

if i put root (/) in url-pattern ..every request to mysite will be filtered ,then in that case how can i get unique id ?

Any solution ?

EDIT:

I dont want to add a directory in url like this :http://www.mysite.com/dynamicpages/708e93

Upvotes: 1

Views: 1160

Answers (1)

Ramesh PVK
Ramesh PVK

Reputation: 15446

For dynamic url's such as use case you have. you should be have all the dynamic pages inside a directory, lets say dynamicpages folder. And you should be mapping a directory url pattern like below :

<filter-mapping>
  <filter-name>foo.bar.MyFilter</filter-name>  
  <url-pattern>/dynmaicpages/*</url-pattern>
</filter-mapping>

<servlet-mapping>
  <servlet-name>foo.bar.MyServlet</servlet-name>  
  <url-pattern>/dynmaicpages/*</url-pattern>
</servlet-mapping>

You can get the Unique id using the following methods:

  • HttpServletRequest.getPathInfo() If you have servlet mapping like above, this will give you the unique id value directly.

For ex, if the request uri is /dynamicpages/708e93 it will return 708e93.

  • HttpServletRequest.getRequestURI() This will give you the entire request url. And you need to parse the unique id properly.

For ex, if the request uri is /dynamicpages/708e93 it will return /dynamicpages/708e93.

Upvotes: 1

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