有一个在线文件(如http://www.example.com/information.asp),我需要抓取并保存到一个目录。我知道有几种逐行抓取和读取在线文件(url)的方法,但是否有一种方法可以使用Java下载并保存文件?


当前回答

总结(并以某种方式润色和更新)之前的答案。以下三种方法实际上是等效的。(我添加了明确的超时,因为我认为这是必须的。没有人希望下载在连接丢失时永远冻结。)

public static void saveUrl1(final Path file, final URL url,
    int secsConnectTimeout, int secsReadTimeout))
    throws MalformedURLException, IOException {

    // Files.createDirectories(file.getParent()); // Optional, make sure parent directory exists
    try (BufferedInputStream in = new BufferedInputStream(
         streamFromUrl(url, secsConnectTimeout,secsReadTimeout));
         OutputStream fout = Files.newOutputStream(file)) {

            final byte data[] = new byte[8192];
            int count;
            while((count = in.read(data)) > 0)
                fout.write(data, 0, count);
        }
}

public static void saveUrl2(final Path file, final URL url,
    int secsConnectTimeout, int secsReadTimeout))
    throws MalformedURLException, IOException {

    // Files.createDirectories(file.getParent()); // Optional, make sure parent directory exists
    try (ReadableByteChannel rbc = Channels.newChannel(
             streamFromUrl(url, secsConnectTimeout, secsReadTimeout)
        );
        FileChannel channel = FileChannel.open(file,
             StandardOpenOption.CREATE,
             StandardOpenOption.TRUNCATE_EXISTING,
             StandardOpenOption.WRITE)
        ) {

        channel.transferFrom(rbc, 0, Long.MAX_VALUE);
    }
}

public static void saveUrl3(final Path file, final URL url,
    int secsConnectTimeout, int secsReadTimeout))
    throws MalformedURLException, IOException {

    // Files.createDirectories(file.getParent()); // Optional, make sure parent directory exists
    try (InputStream in = streamFromUrl(url, secsConnectTimeout,secsReadTimeout) ) {
        Files.copy(in, file, StandardCopyOption.REPLACE_EXISTING);
    }
}

public static InputStream streamFromUrl(URL url,int secsConnectTimeout,int secsReadTimeout) throws IOException {
    URLConnection conn = url.openConnection();
    if(secsConnectTimeout>0)
        conn.setConnectTimeout(secsConnectTimeout*1000);
    if(secsReadTimeout>0)
        conn.setReadTimeout(secsReadTimeout*1000);
    return conn.getInputStream();
}

我没有发现明显的差异,在我看来都是对的。它们既安全又高效。(速度的差异似乎无关紧要——我从本地服务器写入180 MB到SSD磁盘的时间大约在1.2到1.5秒之间波动)。它们不需要外部库。所有这些都可以使用任意大小和(根据我的经验)HTTP重定向。

此外,如果没有找到资源(通常是404错误),所有抛出FileNotFoundException,如果DNS解析失败则抛出java.net.UnknownHostException;其他IOException对应传输过程中的错误。

其他回答

当使用Java 7+时,使用以下方法从Internet下载文件并将其保存到某个目录:

private static Path download(String sourceURL, String targetDirectory) throws IOException
{
    URL url = new URL(sourceURL);
    String fileName = sourceURL.substring(sourceURL.lastIndexOf('/') + 1, sourceURL.length());
    Path targetPath = new File(targetDirectory + File.separator + fileName).toPath();
    Files.copy(url.openStream(), targetPath, StandardCopyOption.REPLACE_EXISTING);

    return targetPath;
}

文档在这里。

下面是用Java代码从网上下载电影的示例代码:

URL url = new
URL("http://103.66.178.220/ftp/HDD2/Hindi%20Movies/2018/Hichki%202018.mkv");
    BufferedInputStream bufferedInputStream = new  BufferedInputStream(url.openStream());
    FileOutputStream stream = new FileOutputStream("/home/sachin/Desktop/test.mkv");

    int count = 0;
    byte[] b1 = new byte[100];

    while((count = bufferedInputStream.read(b1)) != -1) {
        System.out.println("b1:" + b1 + ">>" + count + ">> KB downloaded:" + new File("/home/sachin/Desktop/test.mkv").length()/1024);
        stream.write(b1, 0, count);
    }

试试Java NIO:

URL website = new URL("http://www.website.com/information.asp");
ReadableByteChannel rbc = Channels.newChannel(website.openStream());
FileOutputStream fos = new FileOutputStream("information.html");
fos.getChannel().transferFrom(rbc, 0, Long.MAX_VALUE);

使用transferFrom()可能比从源通道读取并写入此通道的简单循环更有效。许多操作系统可以直接将字节从源通道传输到文件系统缓存中,而不需要实际复制它们。

点击这里查看更多信息。

注意:transferFrom中的第三个参数是传输的最大字节数。整数。MAX_VALUE将传输最多2^31字节,长。MAX_VALUE最多允许2^63字节(比现有的任何文件都大)。

就我个人而言,我发现Apache的HttpClient在这方面比我需要做的任何事情都有能力。这里有一个关于使用HttpClient的很棒的教程

更简单的非阻塞I/O用法:

URL website = new URL("http://www.website.com/information.asp");
try (InputStream in = website.openStream()) {
    Files.copy(in, target, StandardCopyOption.REPLACE_EXISTING);
}