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Java URL encoding of query string parameters

Posted by: admin November 2, 2017 Leave a comment


Say I have a URL


and I have a query entered by the user such as:

random word £500 bank $

I want the result to be a properly encoded URL:


What’s the best way to achieve this? I tried URLEncoder and creating URI/URL objects but none of them come out quite right.


URLEncoder should be 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 = "http://example.com/query?q=" + URLEncoder.encode(q, "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 two encode() methods. One without charset argument and another with. 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.

See also:


I would not use URLEncoder. Besides being incorrectly named (URLEncoder has nothing to do with URLs), inefficient (it uses a StringBuffer instead of Builder and does a couple of other things that are slow) Its also way too easy to screw it up.

Instead I would use URIBuilder or Spring’s org.springframework.web.util.UriUtils.encodeQuery or Commons Apache HttpClient.
The reason being you have to escape the query parameters name (ie BalusC’s answer q) differently than the parameter value.

The only downside to the above (that I found out painfully) is that URL’s are not a true subset of URI’s.

Sample code:

import org.apache.http.client.utils.URIBuilder;

URIBuilder ub = new URIBuilder("http://example.com/query");
ub.addParameter("q", "random word £500 bank $");
String url = ub.toString();

// Result: http://example.com/query?q=random+word+%C2%A3500+bank+%24

Since I’m just linking to other answers I marked this as a community wiki. Feel free to edit.


You need to first create a URI like:

    String urlStr = "http://www.example.com/CEREC® Materials & Accessories/IPS Empress® CAD.pdf"
    URL url= new URL(urlStr);
    URI uri = new URI(url.getProtocol(), url.getUserInfo(), url.getHost(), url.getPort(), url.getPath(), url.getQuery(), url.getRef());

Then convert that Uri to ASCII string:


Now your url string is completely encoded first we did simple url encoding and then we converted it to ASCII String to make sure no character outside US-ASCII are remaining in string. This is exactly how browsers do.


Guava 15 has now added a set of straightforward URL escapers.


Apache Http Components library provides a neat option for building and encoding query params –

With HttpComponents 4.x use –

For HttpClient 3.x use –


Here’s a method you can use in your code to convert a url string and map of parameters to a valid encoded url string containing the query parameters.

String addQueryStringToUrlString(String url, final Map<Object, Object> parameters) throws UnsupportedEncodingException {
    if (parameters == null) {
        return url;

    for (Map.Entry<Object, Object> parameter : parameters.entrySet()) {

        final String encodedKey = URLEncoder.encode(parameter.getKey().toString(), "UTF-8");
        final String encodedValue = URLEncoder.encode(parameter.getValue().toString(), "UTF-8");

        if (!url.contains("?")) {
            url += "?" + encodedKey + "=" + encodedValue;
        } else {
            url += "&" + encodedKey + "=" + encodedValue;

    return url;


I would use this code:

Uri myUI = Uri.parse ("http://example.com/query").buildUpon().appendQueryParameter("q","random word A3500 bank 24").build();

  1. Use this: URLEncoder.encode(query, StandardCharsets.UTF_8.displayName());
    or this:URLEncoder.encode(query, “UTF-8”);
  2. You can use the follwing code.

    String encodedUrl1 = UriUtils.encodeQuery(query, "UTF-8");//not change 
    String encodedUrl2 = URLEncoder.encode(query, "UTF-8");//changed
    String encodedUrl3 = URLEncoder.encode(query, StandardCharsets.UTF_8.displayName());//changed
    System.out.println("url1 " + encodedUrl1 + "\n" + "url2=" + encodedUrl2 + "\n" + "url3=" + encodedUrl3);

You need to first create a URI like This:

var Url="https://http://localhost:50826/#/Post-Detail/12"

var EncodedUri=encodeURIComponent(Url);