Remove one parameter from the query string without using regex

Question

So I have a URL which is

String url = request.getRequestURL()+"?"+request.getQueryString();

There is a parameter Custom="true", which i want to remove before i store it in url. Is there any efficient way to do this, without using regex.

Example - 

 http://myhost:8080/people?lastname=Fox&age=30&Custom=true&verified=yes

 request.getQueryString();  // "lastname=Fox&age=30&Custom=true&verified=yes"

Desired o/p - lastname=Fox&age=30&verified=yes

request.getQueryString().replace("Custom=true&", "") may be? — Eritrean
– Eritrean, Commented Mar 15, 2019 at 10:16

Tarun · Accepted Answer · 2019-03-15 10:14:35Z

1

Just replace Custom=true with empty String

query = query.replace("Custom=true&","");

edited Mar 15, 2019 at 10:14

answered Mar 15, 2019 at 10:10

Tarun

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3 Comments

Shubham Chopra Over a year ago

I guess you did not understand the question, request.getQueryString() already returns string after ?. from that resulting string which is lastname=Fox&age=30&Custom=true&verified=yes" , I want to remove Custom=true

Shashwat Kumar Over a year ago

It will replace otherCustom=true& too

Shubham Chopra Over a year ago

Parameters are unique so that will not be an issue.

Ramesh · Accepted Answer · 2019-03-15 11:11:29Z

0

If you want to replace exact match,below code will work.

String url = "http://myhost:8080/people?lastname=Fox&age=30&Custom=true&verified=yes";
    String newUrl = "";
    String[] splits = url.split("&");
    for(String split : splits) {
        if(!"Custom=true".equals(split)) {
            newUrl = newUrl+"&"+split;
        }
    }

answered Mar 15, 2019 at 11:11

Ramesh

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1 Comment

wheleph Over a year ago

This approach doesn't work if "Custom" is the only parameter in query string

wheleph · Accepted Answer · 2021-02-03 14:39:19Z

0

Here's a solution that relies on Java 8 Streaming API:

    StringBuffer requestUrl = request.getRequestURL();
    String queryString = request.getQueryString();

    String[] queryStringParts = queryString.split("&");

    String newQueryString = Stream.of(queryStringParts)
            .filter(p -> !p.equals("Custom=true"))
            .collect(Collectors.joining("&"));

    String newUrl = StringUtils.hasText(newQueryString) ?
            requestUrl.append("?").append(newQueryString).toString() :
            requestUrl.toString();

Also the fact that request url and query string are initially separate (as stated in the question) actually helps.

answered Feb 3, 2021 at 14:39

wheleph

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