Java从重定向的“友好”网址获取下载文件名

Question

I'm trying to download a file from a given URL which may or may not be a direct link to the file. Does anyone know how I can detect the filename to write to if the URL is an indirect link (ie http://www.example.com/download.php?getFile=1 ) ? It is no problem if the URL is a direct link to extract the filename from the URL and start writing to the extracted filename but with a redirect link the only method I have found so far is to write to an arbitrary filename - foo.txt - and then try and work with that. Problem is I really need the filename (and extension) to be correct. A sample of the code I am using is: (the section in the 'else' clause is neither finished nor working):

public static boolean dlFile(String URL, String dest){
    try{
        URL grab = new URL(URL);
        ReadableByteChannel rbc = Channels.newChannel(grab.openStream());
        String fnRE = ".*/([a-zA-Z0-9\\-\\._]+)$";
            Pattern pattern = Pattern.compile(fnRE);
        Matcher matcher = pattern.matcher(URL);
        String fName = "";
        if(matcher.find()) fName = matcher.group(1);
        else { //filename cannot be extracted - do something here - below doesn't work raises MalformedURLExcpetion
            URL foo = new URL(URL);
            HttpURLConnection fooConnection = (HttpURLConnection) foo.openConnection();
            URL secondFoo = new URL(fooConnection.getHeaderField("Location"));
            System.out.println("Redirect URL: "+secondFoo);
            fooConnection.setInstanceFollowRedirects(false);
            URLConnection fooURL = secondFoo.openConnection();
        }
        System.out.println("Connection to "+URL+" established!");
        if(dest.endsWith("/")){}
        else dest+="/";
        System.out.println("Writing "+fName+" to "+dest);
        FileOutputStream fos = new FileOutputStream(dest+fName);
        fos.getChannel().transferFrom(rbc, 0, 1 << 24);

I am sure there must be a simple way to get the filename from the headers or something like that but I cannot work out how to get it. Thanks in advance,

Answer 1

Assuming the response has a "Location" header field, I was able to obtain the direct link to a url containing multiple redirects like this:

String location = "http://www.example.com/download.php?getFile=1";
HttpURLConnection connection = null;
for (;;) {
    URL url = new URL(location);
    connection = (HttpURLConnection) url.openConnection();
    connection.setInstanceFollowRedirects(false);
    String redirectLocation = connection.getHeaderField("Location");
    if (redirectLocation == null) break;
    location = redirectLocation;
}
//and finally:
String fileName = location.substring(location.lastIndexOf('/') + 1, location.length());

Answer 2

I think its better to use Java Jsoup library, then use the below method:

public static void downloadFileJsoup(String URL, String PATH) throws IOException {
    Response res = Jsoup.connect(URL)
            .userAgent("Mozilla")
            .timeout(30000)
            .followRedirects(true)
            .ignoreContentType(true)
            .maxBodySize(20000000)//Increase value if download is more than 20MB
            .execute(); 
    String remoteFilename=res.header("Content-Disposition").replaceFirst("(?i)^.*filename=\"?([^\"]+)\"?.*$", "$1");
    String filename = PATH + remoteFilename;
    FileOutputStream out = (new FileOutputStream(new java.io.File(filename)));
    out.write( res.bodyAsBytes());
    out.close();
}

Answer 3

No, in general no way. The response does'nt contain that information normally, since you do not add any own protocol information to the data stream (in case you can control the server).

Anyway, you ask for the file name extension. Maybe with the correct content-type you are done.