Replace a Param in QueryString in Java

Question

lets say I have a url param like token=1234235asdjaklj231k209a&name=sam&firname=Mahan how can I replace the value of the token with new one ? I've done something similar to this with pattern and matcher before but I don't recall now but I know there is a way to do so Update : the token can contain any letter but & thanks in advance

@YCF_L the same string but the value of the token is replaced by the new one like token=1234&name=sam&firname=Mahan that 1234 is the new token for instance and the token contains anything but & — Mahan, Sep 19 '17 at 14:30
I don't think this is a job for regex; there are well tested abstractions for urls in java. — Andreas, Sep 19 '17 at 14:39
please explain all the cases you can have so we can help you: — Youcef LAIDANI, Sep 19 '17 at 14:41
possible duplicate of https://stackoverflow.com/questions/13592236/parse-a-uri-string-into-name-value-collection — Andreas, Sep 19 '17 at 14:53

score 10 · Answer 1 · answered Sep 19 '17 at 14:52

10

Spring has a util that handles this need gracefully. Apache httpcomponents does too. Below is a spring example.

import org.springframework.web.util.UriComponentsBuilder;
import java.net.URI;


public class StackOverflow {

  private static class SO46303058 {
    public static void main(String[] args) {
      final String urlString = "https://subdomain.hostname/path/resource?token=1234235asdjaklj231k209a&name=sam&firname=Mahan";
      final URI uri = UriComponentsBuilder.fromHttpUrl(urlString)
          .replaceQueryParam("token", "abc")
          .build().toUri();
      System.out.println(uri);
    }
  }
}

Don't be afraid of adding dependencies to your project, it beats reinventing the wheel.

answered Sep 19 '17 at 14:52

Andreas

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I'm upvoting this because I feel that in general most URI/URL stuff should be handled by such libraries and not manually. – Tim Biegeleisen Sep 19 '17 at 14:53
Note to my downvoter: That being said, if the OP for some reason really only had a fragment of a query string, then such a canned library would not work. – Tim Biegeleisen Sep 19 '17 at 15:04
1

thanks that's great news that I'm using Spring too that this helps a lot and its so clean – Mahan Sep 19 '17 at 15:05

Tim Biegeleisen · Accepted Answer · 2017-09-19T14:52:18.023

3

We can consider doing a simple regex replacement, with a few caveats (q.v. below the code snippet).

String url = "token=1234235asdjaklj231k209a&name=sam&firname=Mahan";
url = url.replaceFirst("\\btoken=.*?(&|$)", "token=new_value$1");
System.out.println(url);
url = "param1=value&token=1234235asdjaklj231k209a";
url = url.replaceFirst("\\btoken=.*?(&|$)", "token=new_value$1");
System.out.println(url);

Edge cases to consider are first that your token may be the last parameter in the query string. To cover this case, we should check for token=... ending in either an ambersand & or the end of the string. But if we don't use a lookahead, and instead consume that ambersand, we have to also add it back in the replacement. The other edge case, correctly caught by @DodgyCodeException in his comment below, is that there be another query parameter which just happens to end in token. To make sure we are really matching our token parameter, we can preface it with a word boundary in the regex, i.e. use \btoken=... to refer to it.

Output:

token=new_value&name=sam&firname=Mahan
param1=value&token=new_value

edited Sep 19 '17 at 14:52

answered Sep 19 '17 at 14:35

Tim Biegeleisen

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What if the query params were "thisisnotatoken=gotcha&token=1234abcd&..." – DodgyCodeException Sep 19 '17 at 14:37
@DodgyCodeException Then we should use a word boundary in this case. You certainly have a nose for dodgy code :-) – Tim Biegeleisen Sep 19 '17 at 14:40
thanks it worked for my cases and that case mentioned above by @DodgyCodeException is not gonna happen for me , but thanks for mentioning that – Mahan Sep 19 '17 at 14:46
1

I'm glad it helped in your specific case, but for general use this is flaky - e.g. there may be a `&` somewhere which will get mistaken by the above. The best way to tackle it would be to use a well-tested URL params parsing library (such as Apache URLEncodedUtils - not saying it's well-tested as after all Apache is a so named because it's a bit patchy, but it's the only one I could find). Then use that to split into a list of name-value pairs, change the appropriate value, and re-encode into a string. – DodgyCodeException Sep 19 '17 at 15:15

score -1 · Answer 3 · answered Sep 19 '17 at 14:34

make a viewModel.

public class veiwModel(){ String token ; // and get and set for exmample }

then use Gson if u have a json text .

   Gson gson = new Gson();
   yourViewModel = gson.fronJson(jsonText , viewModel.class);
   System.out.println(yourViewModel.getToken());

Replace a Param in QueryString in Java

3 Answers3