9

我有一个重定向到另一个 url 的 url。我希望能够获得最终重定向的 URL。我的代码：

    public class testURLConnection
    {
    public static void main(String[] args) throws MalformedURLException, IOException {

    HttpURLConnection con =(HttpURLConnection) new URL( "http://tinyurl.com/KindleWireless" ).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();

} }

它总是给出原始网址，而重定向网址是：http ://www.amazon.com/Kindle-Wireless-Reading-Display-Globally/dp/B003FSUDM4/ref=amb_link_353259562_2?pf_rd_m=ATVPDKIKX0DER&pf_rd_s=center-10&pf_rd_r=11EYKTN682A79T3701&pf_rdp_t= 1270985982&pf_rd_i=B002Y27P3M。

我怎样才能得到这个最终重定向的 URL。

这是我尝试循环直到我们得到重定向。仍然没有获取所需的网址：

    public static String fetchRedirectURL(String url) throws IOException
    {
HttpURLConnection con =(HttpURLConnection) new URL( url ).openConnection();
//System.out.println( "orignal url: " + con.getURL() );
con.setInstanceFollowRedirects(false);
con.connect();


InputStream is = con.getInputStream();
if(con.getResponseCode()==301)
    return con.getHeaderField("Location");
else return null;
    }
    public static void main(String[] args) throws MalformedURLException, IOException {
String url="http://tinyurl.com/KindleWireless";
String fetchedUrl=fetchRedirectURL(url);
System.out.println("FetchedURL is:"+fetchedUrl);
while(fetchedUrl!=null)
{   url=fetchedUrl;
System.out.println("The url is:"+url);
    fetchedUrl=fetchRedirectURL(url);


}
System.out.println(url);

    }
4

7 回答 7

20

试试这个，我使用递归来使用许多重定向 URL。

public static String getFinalURL(String url) throws IOException {
    HttpURLConnection con = (HttpURLConnection) new URL(url).openConnection();
    con.setInstanceFollowRedirects(false);
    con.connect();
    con.getInputStream();

    if (con.getResponseCode() == HttpURLConnection.HTTP_MOVED_PERM || con.getResponseCode() == HttpURLConnection.HTTP_MOVED_TEMP) {
        String redirectUrl = con.getHeaderField("Location");
        return getFinalURL(redirectUrl);
    }
    return url;
}

并使用：

public static void main(String[] args) throws MalformedURLException, IOException {
    String fetchedUrl = getFinalURL("<your_url_here>");
    System.out.println("FetchedURL is:" + fetchedUrl);

}
于 2014-09-25T03:47:24.790 回答
8 
public static String getFinalRedirectedUrl(String url) {

    HttpURLConnection connection;
    String finalUrl = url;
    try {
        do {
            connection = (HttpURLConnection) new URL(finalUrl)
                    .openConnection();
            connection.setInstanceFollowRedirects(false);
            connection.setUseCaches(false);
            connection.setRequestMethod("GET");
            connection.connect();
            int responseCode = connection.getResponseCode();
            if (responseCode >= 300 && responseCode < 400) {
                String redirectedUrl = connection.getHeaderField("Location");
                if (null == redirectedUrl)
                    break;
                finalUrl = redirectedUrl;
                System.out.println("redirected url: " + finalUrl);
            } else
                break;
        } while (connection.getResponseCode() != HttpURLConnection.HTTP_OK);
        connection.disconnect();
    } catch (Exception e) {
        e.printStackTrace();
    }
    return finalUrl;
}
于 2013-09-16T05:43:44.280 回答
2

我的第一个想法是设置instanceFollowRedirects为 false，或者URLConnection改为使用。

在这两种情况下，都不会执行重定向，因此您将收到对原始请求的回复。获取 HTTP 状态值，如果是 3xx，则获取新的重定向值。

当然，可能存在一系列重定向，因此您可能需要迭代直到到达真实（状态 2xx）页面。

于 2013-02-19T07:15:34.027 回答
1

这可能会有所帮助

public static void main(String[] args) throws MalformedURLException,
    IOException {

HttpURLConnection con = (HttpURLConnection) new URL(
        "http://tinyurl.com/KindleWireless").openConnection(proxy);
    System.out.println("orignal url: " + con.getURL());
    con.connect();
    con.setInstanceFollowRedirects(false);
    int responseCode = con.getResponseCode();
    if ((responseCode / 100) == 3) {
        String newLocationHeader = con.getHeaderField("Location");
        responseCode = con.getResponseCode();
        System.out.println("Redirected Location " + newLocationHeader);
        System.out.println(responseCode);
    }

}
于 2013-02-19T09:54:42.980 回答
1

@user719950 在我的 MAC-OSX 上 - 这解决了 HTTP URL 被截断的问题：

在您的原始代码中，只需添加以下行：//您必须通过浏览器查找 IE/Chrome 发送的请求标头是什么。我仍然没有解释为什么这个简单的设置会导致正确的 URL :)

HttpURLConnection con =(HttpURLConnection) new URL
( "http://tinyurl.com/KindleWireless" ).openConnection();
 con.setInstanceFollowRedirects(true);
 con.setDoOutput(true);
  System.out.println( "orignal url: " + con.getURL() );     
         **con.setRequestProperty("User-Agent",
    "Mozilla/5.0 (Macintosh; Intel Mac OS X 10_8_2) 
    AppleWebKit/536.26.17 (KHTML, like Gecko) Version/6.0.2  
   Safari/536.26.17");**                  

           con.connect();
    System.out.println( "connected url: " + con.getURL() );
    Thread.currentThread().sleep(2000l);
    InputStream is = con.getInputStream();
    System.out.println( "redirected url: " + con.getURL() );

    is.close();
于 2013-02-19T09:12:03.387 回答
0

如果有多个重定向，这个会递归：

protected String getDirectUrl(String link) {
    String resultUrl = link;
    HttpURLConnection connection = null;
    try {
        connection = (HttpURLConnection) new URL(link).openConnection();
        connection.setInstanceFollowRedirects(false);
        connection.connect();
        int responseCode = connection.getResponseCode();
        if (responseCode == HttpURLConnection.HTTP_MOVED_PERM || responseCode == HttpURLConnection.HTTP_MOVED_TEMP) {
            String locationUrl = connection.getHeaderField("Location");

            if (locationUrl != null && locationUrl.trim().length() > 0) {
                IOUtils.close(connection);
                resultUrl = getDirectUrl(locationUrl);
            }
        }
    } catch (Exception e) {
        log("error getDirectUrl", e);
    } finally {
        IOUtils.close(connection);
    }
    return resultUrl;
}
于 2014-08-07T19:06:17.547 回答
0

@JEETS您的 fetchRedirectURL 函数可能无法正常工作，因为有多种用于重定向的 HTTP 代码。将其更改为范围检查，它将起作用。

public static String fetchRedirectURL(String url) throws IOException
    {
HttpURLConnection con =(HttpURLConnection) new URL( url ).openConnection();
//System.out.println( "orignal url: " + con.getURL() );
con.setInstanceFollowRedirects(false);
con.connect();

InputStream is = con.getInputStream();
if(con.getResponseCode()>=300 && con.getResponseCode() <400)
    return con.getHeaderField("Location");
else return null;
    }
于 2014-06-20T01:13:28.710 回答