Java - Convert String to valid URI object

Solution 1:

You might try: org.apache.commons.httpclient.util.URIUtil.encodeQuery in Apache commons-httpclient project

Like this (see URIUtil):

URIUtil.encodeQuery("http://www.google.com?q=a b")

will become:

http://www.google.com?q=a%20b

You can of course do it yourself, but URI parsing can get pretty messy...

Solution 2:

Android has always had the Uri class as part of the SDK: http://developer.android.com/reference/android/net/Uri.html

You can simply do something like:

String requestURL = String.format("http://www.example.com/?a=%s&b=%s", Uri.encode("foo bar"), Uri.encode("100% fubar'd"));

Solution 3:

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.

Solution 4:

Even if this is an old post with an already accepted answer, I post my alternative answer because it works well for the present issue and it seems nobody mentioned this method.

With the java.net.URI library:

URI uri = URI.create(URLString);

And if you want a URL-formatted string corresponding to it:

String validURLString = uri.toASCIIString();

Unlike many other methods (e.g. java.net.URLEncoder) this one replaces only unsafe ASCII characters (like ç, é...).

In the above example, if URLString is the following String:

"http://www.domain.com/façon+word"

the resulting validURLString will be:

"http://www.domain.com/fa%C3%A7on+word"

which is a well-formatted URL.

Solution 5:

If you don't like libraries, how about this?

Note that you should not use this function on the whole URL, instead you should use this on the components...e.g. just the "a b" component, as you build up the URL - otherwise the computer won't know what characters are supposed to have a special meaning and which ones are supposed to have a literal meaning.

/** Converts a string into something you can safely insert into a URL. */
public static String encodeURIcomponent(String s)
{
    StringBuilder o = new StringBuilder();
    for (char ch : s.toCharArray()) {
        if (isUnsafe(ch)) {
            o.append('%');
            o.append(toHex(ch / 16));
            o.append(toHex(ch % 16));
        }
        else o.append(ch);
    }
    return o.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;
}