Remove request parameter from query string

Question

I have a query string that could be:

/fr/hello?language=en

or

/fr/welcome?param1=222&param2=aloa&language=en

or

/it/welcome?param1=222&language=en&param2=aa

I would like to remove from each query string the parameter language with its value, therefore the results would be:

/fr/hello

and

/fr/welcome?param1=222&param2=aloa

and

/it/welcome?param1=222&param2=aa

EDIT: The length of the value of the parameter could be more than 2

Does anybody know any good regex expression to use in String.replaceAll([regex],[replace]) ?

no, it doesn't work in my case, obviously I have replace 'foo' with 'language' — cloudy_weather
– cloudy_weather, Commented Jul 16, 2014 at 11:52

Avinash Raj · Accepted Answer · 2014-07-16 12:39:39Z

5

Use the below regex and replace the matched strings with empty string,

[&?]language.*?(?=&|\?|$)

DEMO

Example code:

String s1 = "/fr/welcome?param1=222&param2=aloa&language=en";
String s2 = "/fr/welcome?language=en";
String s3 = "/fr/welcome?param1=222&language=en&param2=aa";
String m1 = s1.replaceAll("[&?]language.*?(?=&|\\?|$)", "");
String m2 = s2.replaceAll("[&?]language.*?(?=&|\\?|$)", "");
String m3 = s3.replaceAll("[&?]language.*?(?=&|\\?|$)", "");
System.out.println(m1);
System.out.println(m2);
System.out.println(m3);

Output:

/fr/welcome?param1=222&param2=aloa
/fr/welcome
/fr/welcome?param1=222&param2=aa

IDEONE 1 or IDEONE 2

edited Jul 16, 2014 at 12:39

answered Jul 16, 2014 at 11:54

Avinash Raj

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8 Comments

cloudy_weather Over a year ago

I can't understand why I doesn't match in my code: queryString.replaceAll("[&?]language.*?(?=&|\\?|$)", "");

cloudy_weather Over a year ago

but it does not work for this: /en/welcome?language=en

Avinash Raj Over a year ago

@daniele now see the ideone link.

Alexander Klimetschek Over a year ago

Does not work for /it/welcome?language=en&param1=222 which ends up as /it/welcome&param1=222 (query does not start with ? anymore). As a solution, one can split of the query string after the ? in a first step, then use [&]language.*?(?=&|\?|$) for the replacement on the query string, and later build the full URL again using path + "?" + replacedQuery.

Vigneshwaran Over a year ago

@AvinashRaj for input /fr/welcome?language=en&a=b gives /fr/welcome&a=b

|

TheLostMind · Accepted Answer · 2014-07-16 12:03:15Z

5

You could use regex with replaceAll()

public static void main(String[] args) {
    String s1 = "/fr/welcome?language=en";
    String s2 = "/fr/welcome?param1=222&param2=aloa&language=en";
    String s3 = "/fr/welcome?param1=222&language=en&param2=aa";
    String pattern = "[?&]language=.{2}"; // use pattern = "([?&]language=\\w+)"; for more than 2 letters after language ==.
    System.out.println(s1.replaceAll(pattern, ""));
    System.out.println(s2.replaceAll(pattern, ""));
    System.out.println(s3.replaceAll(pattern, ""));
}

o/p :

/fr/welcome
/fr/welcome?param1=222&param2=aloa
/fr/welcome?param1=222&param2=aa

edited Jul 16, 2014 at 12:03

answered Jul 16, 2014 at 11:54

TheLostMind

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1 Comment

M. Justin Over a year ago

This messes up the URL if there are more than one query parameters and the first one is "language". In this case, the "?" is removed, so the query parameters are no longer treated as query parameters, but instead as part of the URL path.

Jens · Accepted Answer · 2014-07-16 11:50:58Z

1

This regexp should help you:

"language=\\w{2}"

answered Jul 16, 2014 at 11:50

Jens

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1 Comment

cloudy_weather Over a year ago

the length is not necessarily 2

Akash Thakare · Accepted Answer · 2014-07-16 12:03:37Z

1

I would like to remove from each query string the parameter language with its value,...

You can use replaceAll.

String s="/fr/welcome?language=en";
s=s.replaceAll("(\\?|&)language=\\w+", "");

(\\?|&) group will match ? or &
\\w+ will match one or more word character

edited Jul 16, 2014 at 12:03

answered Jul 16, 2014 at 11:57

Akash Thakare

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1 Comment

M. Justin Over a year ago

This messes up the URL if there are more than one query parameters and the first one is "language". In this case, the "?" is removed, so the query parameters are no longer treated as query parameters, but instead as part of the URL path.

Bahadir Tasdemir · Accepted Answer · 2017-09-18 12:32:40Z

1

This will remove any parameter properly, even if it is placed more than one (for example="/fr/welcome?language=en&param1=222&param2=aloa")

public String removeParamFromUrl(final String url, final String param) {
    if (StringUtils.isNotBlank(url)) {
        return url.replaceAll("&" + param + "=[^&]+", "")
                .replaceAll("\\?" + param + "=[^&]+&", "?")
                .replaceAll("\\?" + param + "=[^&]+", "");
    } else {
        return url;
    }
}

answered Sep 18, 2017 at 12:32

Bahadir Tasdemir

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Comments

M. Justin · Accepted Answer · 2021-05-24 18:35:00Z

Rather than using a regex, it may be better to use a dedicated URI-manipulation API to remove the query parameter. The Spring UriComponentsBuilder class can be used to remove the given query parameter, retaining the rest. I'm assuming a Spring-specific solution is acceptable, as this question is tagged with spring.

private static String removeQueryParam(String url) {
    return UriComponentsBuilder.fromUriString(url)
            .replaceQueryParam("language")
            .build()
            .toUriString();
}

From the question as asked, it's unclear why or whether a regex-based solution using String.replaceAll is necessary, or whether instead any Java or Spring-based solution would be acceptable. In other words, this may be an XY problem where the goal is to remove the "language" query parameter while retaining all other query parameters, and there's no particular reason a regex needs to be involved in the solution.

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