CODEWITHSUNDEEP
javaandroidhttphttp-redirecthttpurlconnection
Julian
Julian

Reputation: 2101

Android HttpURLConnection: Handle HTTP redirects

I'm using HttpURLConnection to retrieve an URL just like that:

URL url = new URL(address);
HttpURLConnection connection = (HttpURLConnection) url.openConnection();
connection.setInstanceFollowRedirects(true);
// ...

I now want to find out if there was a redirect and if it was a permanent (301) or temporary (302) one in order to update the URL in the database in the first case but not in the second one.

Is this possible while still using the redirect handling of HttpURLConnection and if, how?

Upvotes: 12

Views: 16705

Answers (2)

danik
danik

Reputation: 856

private HttpURLConnection openConnection(String url) throws IOException {
    HttpURLConnection connection;
    boolean redirected;
    do {
        connection = (HttpURLConnection) new URL(url).openConnection();
        int code = connection.getResponseCode();
        redirected = code == HTTP_MOVED_PERM || code == HTTP_MOVED_TEMP || code == HTTP_SEE_OTHER;
        if (redirected) {
            url = connection.getHeaderField("Location");
            connection.disconnect();
        }
    } while (redirected);
    return connection;
}

Upvotes: 9

syb0rg
syb0rg

Reputation: 8247

Simply call getUrl() on URLConnection instance after calling getInputStream():

URLConnection con = new URL(url).openConnection();
System.out.println("Orignal URL: " + con.getURL());
con.connect();
System.out.println("Connected URL: " + con.getURL());
InputStream is = con.getInputStream();
System.out.println("Redirected URL: " + con.getURL());
is.close();

If you need to know whether the redirection happened before actually getting it's contents, here is the sample code:

HttpURLConnection con = (HttpURLConnection) (new URL(url).openConnection());
con.setInstanceFollowRedirects(false);
con.connect();
int responseCode = con.getResponseCode();
System.out.println(responseCode);
String location = con.getHeaderField("Location");
System.out.println(location);

Upvotes: 11

Related Questions