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.

Solution 1

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 = "" + 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.

See also:

Solution 2

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("");
ub.addParameter("q", "random word £500 bank \$");
String url = ub.toString();

// Result:

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

Solution 3

You need to first create a URI like:

String urlStr = "® 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.

Solution 4

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

Solution 5

URL url= new URL(" word £500 bank $");
URI uri = new URI(url.getProtocol(), url.getUserInfo(), IDN.toASCII(url.getHost()), url.getPort(), url.getPath(), url.getQuery(), url.getRef());
String correctEncodedURL=uri.toASCIIString(); 


What is happening here?

1. Split URL into structural parts. Use for it.

2. Encode each structural part properly!

3. Use IDN.toASCII(putDomainNameHere) to Punycode encode the host name!

4. Use to percent-encode, NFC encoded unicode - (better would be NFKC!). For more info see: How to encode properly this URL

In some cases it is advisable to check if the url is already encoded. Also replace '+' encoded spaces with '%20' encoded spaces.

Here are some examples that will also work properly

      "in" : "http://نامهای.com/",
     "out" : ""
     "in" : "",
     "out" : ""
     "in" : " book.pdf", 
     "out" : ""
}, {
     "in" : " word £500 bank $", 
     "out" : "$"

The solution passes around 100 of the testcases provided by Web Plattform Tests.

Solution 6

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

With HttpComponents 4.x use - URLEncodedUtils

For HttpClient 3.x use - EncodingUtil

Solution 7

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;

Solution 8

Using Spring's UriComponentsBuilder:


Solution 9

In android I would use this code:

Uri myUI = Uri.parse ("").buildUpon().appendQueryParameter("q","random word A3500 bank 24").build();

Where Uri is a

Solution 10

In my case i just needed to pass the whole url and encode only the value of each parameters. I didn't find a common code to do that so (!!) so i created this small method to do the job :

public static String encodeUrl(String url) throws Exception {
    if (url == null || !url.contains("?")) {
        return url;

    List<String> list = new ArrayList<>();
    String rootUrl = url.split("\\?")[0] + "?";
    String paramsUrl = url.replace(rootUrl, "");
    List<String> paramsUrlList = Arrays.asList(paramsUrl.split("&"));
    for (String param : paramsUrlList) {
        if (param.contains("=")) {
            String key = param.split("=")[0];
            String value = param.replace(key + "=", "");
            list.add(key + "=" +  URLEncoder.encode(value, "UTF-8"));
        else {

    return rootUrl + StringUtils.join(list, "&");

public static String decodeUrl(String url) throws Exception {
    return URLDecoder.decode(url, "UTF-8");

It uses org.apache.commons.lang3.StringUtils

Solution 11

  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);