Parsing and encoding URL before sending to Rest webservice

Question

I am developing an App where I am sending get request to REST web service using @RequestMapping(value = "myURLpattern", method = RequestMethod.GET, produces = "image/png")

I came across a bug if somebody enters # in the URL query String it breaks from there http://myserver.com:8080/csg?x=1&y=2#4&z=32&p=1

So I tried to encode it via filter HTTPRequestWrapper so that # will be replaced by %23 using URLEncoder.

My problem is for encoding URL, first I need to read the URL(request.getRequestURL()). and getURL again can't read string after #. request.getRequestURL() will return only (http://myserver.com:8080/csg?x=1&y=2)

Is there any other way to parse the complete url and encode it before sending to REST web service?

Community · Accepted Answer · 2017-05-23 12:16:04Z

3

That is not a bug. Check this:

What is the meaning of # in URL and how can i use that?

A # in the url query string is wrong. You should encode it on client side, before sending it to your server. See this one: How to pass '#' in query string It is in asp, but you should find the java equivalent.

Something to get you started can be this one Java URL encoding of query string parameters

edited May 23, 2017 at 12:16

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answered Sep 6, 2013 at 12:44

Nikola Yovchev

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2 Comments

Krishna Over a year ago

Thanks for your reference so if some one enters # in one of Html field like serial number:12#ser4 then we can not parse it on server side. myserver.com:8080/… for the above scenario how we are going to handle it

Nikola Yovchev Over a year ago

Encode it on the client side. Take a look at encodeURIComponent function in javascriopt. encodeURIComponent('12#ser4'); would make '12#ser4' become '12%23ser4' which is what you should be sending to the server, so your request would look like that: myserver.com:8080/…

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