Encoding Url Query Parameters in Java

Encoding URL query parameters in Java

java.net.URLEncoder.encode(String s, String encoding) can help too. It follows the HTML form encoding application/x-www-form-urlencoded.

URLEncoder.encode(query, "UTF-8");

On the other hand, Percent-encoding (also known as URL encoding) encodes space with %20. Colon is a reserved character, so : will still remain a colon, after encoding.

Encoding a URL Query Parameter so it can have a '+'

Found what I believe to be a decent solution. It turns out that a large part of the problem is actually the "exchange" function, which takes a string for a URL, but then re-encodes that URL for reasons I cannot fathom. However, the exchange function can be sent a java.net.URI instead. In this case, it does not try to interpolate anything, as it is already a URI. I then use java.net.URLEncoder.encode() to encode the pieces. I still have no idea why this isn't standard in Spring, but this should work.

    private String mapToQueryString(Map<String, String> query) {
        List<String> entries = new LinkedList<String>();
        for (Map.Entry<String, String> entry : query.entrySet()) {
            try {
                entries.add(URLEncoder.encode(entry.getKey(), "UTF-8") + "=" + URLEncoder.encode(entry.getValue(), "UTF-8"));
            } catch(Exception e) {
                log.error("Unable to encode string for URL: " + entry.getKey() + " / " + entry.getValue(), e);
            }
        }
        return String.join("&", entries);
    }

    /* Later in the code */
    String endpoint = "https://baseurl.example.com/blah";
    String finalUrl = query.isEmpty() ? endpoint : endpoint + "?" + mapToQueryString(query);
    URI uri;
    try {
        uri = new URI(finalUrl);
    } catch(URISyntaxException e) {
        log.error("Bad URL // " + finalUrl, e);
            return null;
        }
    }
    /* ... */
    HttpEntity<TheResponse> resp = myRestTemplate.exchange(uri, ...)

Java URL encoding of query string parameters

URLEncoder is the way to go. You only need to keep in mind to encode only the individual query string parameter name and/or value, not the entire URL, for sure not the query string parameter separator character & nor the parameter name-value separator character =.

String q = "random word £500 bank $";
String url = "https://example.com?q=" + URLEncoder.encode(q, StandardCharsets.UTF_8);

When you're still not on Java 10 or newer, then use StandardCharsets.UTF_8.toString() as charset argument, or when you're still not on Java 7 or newer, then use "UTF-8".

Note that spaces in query parameters are represented by +, not %20, which is legitimately valid. The %20 is usually to be used to represent spaces in URI itself (the part before the URI-query string separator character ?), not in query string (the part after ?).

Also note that there are three encode() methods. One without Charset as second argument and another with String as second argument which throws a checked exception. The one without Charset argument is deprecated. Never use it and always specify the Charset argument. The javadoc even explicitly recommends to use the UTF-8 encoding, as mandated by RFC3986 and W3C.

All other characters are unsafe and are first converted into one or more bytes using some encoding scheme. Then each byte is represented by the 3-character string "%xy", where xy is the two-digit hexadecimal representation of the byte. The recommended encoding scheme to use is UTF-8. However, for compatibility reasons, if an encoding is not specified, then the default encoding of the platform is used.

Encode & in query parameter value in string url in java

Here is how I have solved my problem

private URI convertStringURLToURI(String url) {
    URI uri = null;

    UriComponentsBuilder uriComponentsBuilder = UriComponentsBuilder.fromUriString(url);
    String queryParameter = uriComponentsBuilder.build().toUri().getQuery();

    if(queryParameter != null) {

        // Split the query part into queries
        String[] splitedQueryParameter = queryParameter.split("&");

        Stack<String> queryParamStack = new Stack<>();

        for(int i = 0; i < splitedQueryParameter.length; i++) {
            /*
                In case "&" is present in any of the query values ex. b=abc & xyz then
                it would have split as "b=abc" and "xyz" due to "&" symbol

                Below code handle such situation
                If "=" is present in any value then it is pushed to stack,
                else "=" is not present then this value was part of the previous push query
                due to "&" present in query value, so the else part handles this situation
             */
            if(splitedQueryParameter[i].contains("=")) {
                queryParamStack.push(splitedQueryParameter[i]);
            } else {
                String oldValue = queryParamStack.pop();
                String newValue = oldValue + "&" + splitedQueryParameter[i];
                queryParamStack.push(newValue);
            }
        }

        MultiValueMap<String, String> queryParams = new LinkedMultiValueMap<>();
        while(!queryParamStack.isEmpty()) {
            String[] query = queryParamStack.pop().split("=");
            String queryParameterValue = query[1];
        /*
            If in the query value, "=" is present somewhere then the query value would have
            split into more than two parts, so below for loop handles that situation
         */
            for(int i = 2; i < query.length; i++) {
                queryParameterValue = queryParameterValue + "=" + query[i];
            }
            // encoding the query value and adding to Map
            queryParams.add(query[0], UriUtils.encode(queryParameterValue, "UTF-8"));
        }

        uriComponentsBuilder.replaceQueryParams(queryParams);

        try {
            uri = new URI(uriComponentsBuilder.build().toString());
        } catch (URISyntaxException e) {
            e.printStackTrace();
        }

    }
    return uri;
}

So, this String URL

String url = "http://www.example.com/site?a=abc&b=qwe & asd&c=foo ?bar=2";

becomes

URI uri = "http://www.example.com/site?c=foo%20%3Fbar%3D2&b=qwe%20%26%20asd&a=abc"

Here, the sequence of query gets changed due to the use of Stack, it will not cause any issue in executing the uri, however you can handle that by modifying the code

Encode URL parameters using Java

This answer has a good discussion of encoding the various parts of a URI/URL. You're on the right track, but your specific problem is that you have the various parts of the URI wrong. You need to use the multi-part constructor that takes an authority, path, query, and fragment:

URI uri = new URI("http", "statebuild-dev.com", "/iit-title/size.xml", "id=(102 or 104)", null);
System.out.println(uri.toString());
System.out.println(uri.toASCIIString());

Encoding query parameters in rest assured

You need to provide a second parameter to give the character-set to use for the encoding to take place:

String restUrl = URLEncoder.encode(queryParam, StandardCharsets.UTF_8);

Should work

https://docs.oracle.com/javase/7/docs/api/java/net/URLEncoder.html Second method is listed below the deprecated one you're using.

Encoding Url Query Parameters in Java