HTTP URL Address Encoding in Java

a solution i developed and much more stable than any other:

public class URLParamEncoder {

    public static String encode(String input) {
        StringBuilder resultStr = new StringBuilder();
        for (char ch : input.toCharArray()) {
            if (isUnsafe(ch)) {
                resultStr.append('%');
                resultStr.append(toHex(ch / 16));
                resultStr.append(toHex(ch % 16));
            } else {
                resultStr.append(ch);
            }
        }
        return resultStr.toString();
    }

    private static char toHex(int ch) {
        return (char) (ch < 10 ? '0' + ch : 'A' + ch - 10);
    }

    private static boolean isUnsafe(char ch) {
        if (ch > 128 || ch < 0)
            return true;
        return " %$&+,/:;=?@<>#%".indexOf(ch) >= 0;
    }

}

The java.net.URI class can help; in the documentation of URL you find

Note, the URI class does perform escaping of its component fields in certain circumstances. The recommended way to manage the encoding and decoding of URLs is to use an URI

Use one of the constructors with more than one argument, like:

URI uri = new URI(
    "http", 
    "search.barnesandnoble.com", 
    "/booksearch/first book.pdf",
    null);
URL url = uri.toURL();
//or String request = uri.toString();

(the single-argument constructor of URI does NOT escape illegal characters)

---

Only illegal characters get escaped by above code - it does NOT escape non-ASCII characters (see fatih's comment).
The toASCIIString method can be used to get a String only with US-ASCII characters:

URI uri = new URI(
    "http", 
    "search.barnesandnoble.com", 
    "/booksearch/é",
    null);
String request = uri.toASCIIString();

---

For an URL with a query like http://www.google.com/ig/api?weather=São Paulo, use the 5-parameter version of the constructor:

URI uri = new URI(
        "http", 
        "www.google.com", 
        "/ig/api",
        "weather=São Paulo",
        null);
String request = uri.toASCIIString();

I'm going to add one suggestion here aimed at Android users. You can do this which avoids having to get any external libraries. Also, all the search/replace characters solutions suggested in some of the answers above are perilous and should be avoided.

Give this a try:

String urlStr = "http://abc.dev.domain.com/0007AC/ads/800x480 15sec h.264.mp4";
URL url = new URL(urlStr);
URI uri = new URI(url.getProtocol(), url.getUserInfo(), url.getHost(), url.getPort(), url.getPath(), url.getQuery(), url.getRef());
url = uri.toURL();

You can see that in this particular URL, I need to have those spaces encoded so that I can use it for a request.

This takes advantage of a couple features available to you in Android classes. First, the URL class can break a url into its proper components so there is no need for you to do any string search/replace work. Secondly, this approach takes advantage of the URI class feature of properly escaping components when you construct a URI via components rather than from a single string.

The beauty of this approach is that you can take any valid url string and have it work without needing any special knowledge of it yourself.

Please be warned that most of the answers above are INCORRECT.

The URLEncoder class, despite is name, is NOT what needs to be here. It's unfortunate that Sun named this class so annoyingly. URLEncoder is meant for passing data as parameters, not for encoding the URL itself.

In other words, "http://search.barnesandnoble.com/booksearch/first book.pdf" is the URL. Parameters would be, for example, "http://search.barnesandnoble.com/booksearch/first book.pdf?parameter1=this&param2=that". The parameters are what you would use URLEncoder for.

The following two examples highlights the differences between the two.

The following produces the wrong parameters, according to the HTTP standard. Note the ampersand (&) and plus (+) are encoded incorrectly.

uri = new URI("http", null, "www.google.com", 80, 
"/help/me/book name+me/", "MY CRZY QUERY! +&+ :)", null);

// URI: http://www.google.com:80/help/me/book%20name+me/?MY%20CRZY%20QUERY!%20+&+%20:)

The following will produce the correct parameters, with the query properly encoded. Note the spaces, ampersands, and plus marks.

uri = new URI("http", null, "www.google.com", 80, "/help/me/book name+me/", URLEncoder.encode("MY CRZY QUERY! +&+ :)", "UTF-8"), null);

// URI: http://www.google.com:80/help/me/book%20name+me/?MY+CRZY+QUERY%2521+%252B%2526%252B+%253A%2529