URL encoding in Android

问题

How do you encode a URL in Android?

I thought it was like this:

final String encodedURL = URLEncoder.encode(urlAsString, \"UTF-8\");
URL url = new URL(encodedURL);

If I do the above, the http:// in urlAsString is replaced by http%3A%2F%2F in encodedURL and then I get a java.net.MalformedURLException when I use the URL.

回答1:

You don't encode the entire URL, only parts of it that come from "unreliable sources".

String query = URLEncoder.encode("apples oranges", "utf-8");
String url = "http://stackoverflow.com/search?q=" + query;

Alternatively, you can use Strings.urlEncode(String str) of DroidParts that doesn't throw checked exceptions.

Or use something like

String uri = Uri.parse("http://...")
                .buildUpon()
                .appendQueryParameter("key", "val")
                .build().toString();

回答2:

I'm going to add one suggestion here. You can do this which avoids having to get any external libraries.

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.

回答3:

For android, I would use String android.net.Uri.encode(String s)

Encodes characters in the given string as '%'-escaped octets using the UTF-8 scheme. Leaves letters ("A-Z", "a-z"), numbers ("0-9"), and unreserved characters ("_-!.~'()*") intact. Encodes all other characters.

Ex/

String urlEncoded = "http://stackoverflow.com/search?q=" + Uri.encode(query);

回答4:

Also you can use this

private static final String ALLOWED_URI_CHARS = "@#&=*+-_.,:!?()/~'%";
String urlEncoded = Uri.encode(path, ALLOWED_URI_CHARS);

it's the most simple method

回答5:

try {
                    query = URLEncoder.encode(query, "utf-8");
                } catch (UnsupportedEncodingException e) {
                    // TODO Auto-generated catch block
                    e.printStackTrace();
                }

回答6:

you can use below methods

public static String parseUrl(String surl) throws Exception
{
    URL u = new URL(surl);
    return new URI(u.getProtocol(), u.getAuthority(), u.getPath(), u.getQuery(), u.getRef()).toString();
}

public String parseURL(String url, Map<String, String> params)
{
    Builder builder = Uri.parse(url).buildUpon();
    for (String key : params.keySet())
    {
        builder.appendQueryParameter(key, params.get(key));
    }
    return builder.build().toString();
}

the second one is better than first.

来源：https://stackoverflow.com/questions/3286067/url-encoding-in-android

标签

android

url

urlencode