[java] Java URL encoding of query string parameters

Say I have a URL

http://example.com/query?q=

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:

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.

Using Spring's UriComponentsBuilder:

UriComponentsBuilder
        .fromUriString(url)
        .build()
        .encode()
        .toUri()

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

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);
   

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;
}

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

I found an easy solution to your question. I also wanted to use an encoded URL but nothing helped me.

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

to use String example = "random word £500 bank $"; you can you below code.

String example = "random word £500 bank $";
String URL = "http://example.com/query?q=" + example.replaceAll(" ","%20");

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.

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

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

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.

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$

What is happening here?

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

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

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