Question

I am looking to parse a URL to obtain a collection of the querystring parameters in Java. To be clear, I need to parse a given URL(or string value of a URL object), not the URL from a servlet request.

It looks as if the javax.servlet.http.HttpUtils.parseQueryString method would be the obvious choice, but it has been deprecated.

Is there an alternative method that I am missing, or has it just been deprecated without an equivalent replacement/enhanced function?

Was it helpful?

Solution

Well, as you mention that the URL does not come from a servlet request, the right answer is, as usual, it depends.

The problem with query part of an url is that there is no clear specification about how to handle parameters duplication.

For example, consider an url like this one:

http://www.example.com?param1=value1&param2=value2&param1=value3

What do you expect as a value for param1? the first value, the last one, an array? The issue is that, according to the specs, all these answers are valid and server vendor are free to support one of these or another. Some use the param1[] notation to indicate that it has to be treated as an array, but again, this is not a unified solution.

So the "best" solution is to know how your destination handle parameters, and mimic the behaviour with a self-made utility class.

OTHER TIPS

I think the idea is to use the HttpServletRequest instead. There is the getParameterMap(), getParameterNames() and getParameterValues() methods to start.

There is also the getParameter(String paramname) method to get the value of a specific method.

These make no distinction between querystring parameters and form parameters though so if your intention was to look for a querystring in aparticular then I guess this wouldn't help.

As far as I know, there isn't one.

It shouldn't be too difficult to write one yourself though. The hardest part, I imagine, would be decoding the URL name/values (which really isn't that hard, if you think about it), and you can use java.net.URLDecoder#decodeURL(String,String) for that.

Licensed under: CC-BY-SA with attribution
Not affiliated with StackOverflow
scroll top