I want the result to be a properly encoded URL:

http://example.com/query?q=random%20word%20%A3500%20bank%20%24

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

I have used URI.create and replaced spaces with + in querystring. At the client site it converted + back to spaces when I selected the query strings. That has worked for me. – ND27 Jun 17 '14 at 16:31 Possible duplicate of Is there a Java method that encodes a collection of parameters as a URL query component? – Nick Grealy Apr 13 '17 at 5:07

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:

  • What every web developer must know about URL encoding
  • @sharadendusinha: as documented and answered, URLEncoder is for URL-encoded query parameters conform application/x-www-form-urlencoded rules. Path parameters don't fit in this category. You need an URI encoder instead. – BalusC Jul 30 '17 at 13:18 Adding to @TmTron 's comment, more precisely one should use the string constant StandardCharsets.UTF_8.name() – Jose Duarte Feb 28 '18 at 17:48 @WijaySharma: Because URL-specific characters would get encoded as well. You should only do that when you want to pass the entire URL as a query parameter of another URL. – BalusC Mar 15 '18 at 9:59 In Java 10 the method URLEncoder.encode​(String, Charset) has been added. With this method you do not have to deal with impossible UnsupportedEncodingExceptions anymore. – Marcono1234 Aug 1 at 15:45

    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.

    @Luis: URLEncoder is as its javadoc says intented to encode query string parameters conform application/x-www-form-urlencoded as described in HTML spec: w3.org/TR/html4/interact/…. Some users indeed confuse/abuse it for encoding whole URIs, like the current answerer apparently did. – BalusC Feb 3 '15 at 18:15 @LuisSep in short URLEncoder is for encoding for form submission. It is not for escaping. Its not the exact same escaping that you would use to create URLs to be put in your web page but happens to be similar enough that people abuse it. The only time you should be using URLEncoder is if your writing a HTTP client (and even then there are far superior options for encoding). – Adam Gent Feb 3 '15 at 19:48 @BalusC "Some users indeed confuse/abuse it for encoding whole URIs, like the current answerer apparently did.". You assumed wrong. I never said I screwed up with it. I have just seen others that have done it, who's bugs I have to fix. The part that I screwed up is that the Java URL class will accept unescaped brackets but not the URI class. There are a lot of way to screw up constructing URLs and not everyone is brilliant like you. I would say that most users that are looking on SO for URLEncoding probably are "users indeed confuse/abuse" URI escaping. – Adam Gent Feb 3 '15 at 20:12

    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:

        urlStr=uri.toASCIIString();
    

    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.

    Thanks! It's stupid that your solution works, but built-in URL.toURI() doesn't. – user11153 Mar 25 '15 at 12:45 Unfortunately this doesn't seem to work with "file:///" (e.g.: "file:///some/directory/a file containing spaces.html"); it bombs with MalformedURLException in "new URL()"; any idea how to fix this? – ZioByte Apr 30 '15 at 10:23 You need to do something like this: String urlStr = "some/directory/a file containing spaces.html"; URL url= new URL(urlStr); URI uri = new URI(url.getProtocol(), url.getUserInfo(), url.getHost(), url.getPort(), url.getPath(), url.getQuery(), url.getRef()); urlStr=uri.toASCIIString(); urlStr.replace("http://","file:///"); I have not tested it, but I think it will work.... :) – M Abdul Sami Apr 30 '15 at 20:14 @tibi you can simply use uri.toString() method to convert it to string instead of Ascii string. – M Abdul Sami Sep 9 '15 at 3:40 The API I was working with didn't accept the + replacement for spaces, but accepted the %20 so this solution worked better than BalusC, thanks! – Julian Honma Sep 1 '17 at 12:44 not sure they have the problem. they differentiate for instance "+" or "%20" to escape " " (form param or path param) which URLEncoder doesn't. – Emmanuel Touzery Apr 16 '15 at 11:01 This worked for me I just replaced call to URLEncoder() to call to UrlEscapers.urlFragmentEscaper() and it worked, not clear if I should be using UrlEscapers.urlPathSegmentEscaper() instead. – Paul Taylor Nov 2 '15 at 12:18 Actually it didnt work for me because unlike URLEncoder it doesnt encode '+' it leaves it alone, server decodes '+' as space whereas if I use URLEncoder '+'s are converted to %2B and correctly decoded back to + – Paul Taylor Nov 2 '15 at 17:52

    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

    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;
    

    Use the following standard Java solution (passes around 100 of the testcases provided by Web Plattform Tests):

    0. Test if URL is already encoded. Replace '+' encoded spaces with '%20' encoded spaces.

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

    2. Encode each structural part properly!

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

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

    URL url= new URL("http://example.com/query?q=random 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(); 
    System.out.println(correctEncodedURL);
    

    Prints

    http://example.com/query?q=random%20word%20%C2%A3500%20bank%20$
    

    Here are some examples that will also work properly

    "in" : "http://نامه‌ای.com/", "out" : "http://xn--mgba3gch31f.com/" "in" : "http://www.example.com/‥/foo", "out" : "http://www.example.com/%E2%80%A5/foo" "in" : "http://search.barnesandnoble.com/booksearch/first book.pdf", "out" : "http://search.barnesandnoble.com/booksearch/first%20book.pdf" "in" : "http://example.com/query?q=random word £500 bank $", "out" : "http://example.com/query?q=random%20word%20%C2%A3500%20bank%20$"

    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 {
                list.add(param);
        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

    In android I would use this code:

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

    Where Uri is a android.net.Uri

  • Use this: URLEncoder.encode(query, StandardCharsets.UTF_8.displayName()); or this:URLEncoder.encode(query, "UTF-8");
  • 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);
                    Not correct. You have to encode the parameter names and values separately. Encoding the entire query string will also encode the = and & separators, which is not correct.
                        – user207421
                    Feb 13 '18 at 1:40
                            
                                
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