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

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

其他回答

可以使用Apache的HttpComponents而不是Commons IO来下载文件。这段代码允许您根据URL在Java中下载文件,并将其保存到特定的目的地。

public static boolean saveFile(URL fileURL, String fileSavePath) {

    boolean isSucceed = true;

    CloseableHttpClient httpClient = HttpClients.createDefault();

    HttpGet httpGet = new HttpGet(fileURL.toString());
    httpGet.addHeader("User-Agent", "Mozilla/5.0 (Windows NT 6.3; WOW64; rv:34.0) Gecko/20100101 Firefox/34.0");
    httpGet.addHeader("Referer", "https://www.google.com");

    try {
        CloseableHttpResponse httpResponse = httpClient.execute(httpGet);
        HttpEntity fileEntity = httpResponse.getEntity();

        if (fileEntity != null) {
            FileUtils.copyInputStreamToFile(fileEntity.getContent(), new File(fileSavePath));
        }

    } catch (IOException e) {
        isSucceed = false;
    }

    httpGet.releaseConnection();

    return isSucceed;
}

与单行代码相比:

FileUtils.copyURLToFile(fileURL, new File(fileSavePath),
                        URLS_FETCH_TIMEOUT, URLS_FETCH_TIMEOUT);

这段代码将使您对进程有更多的控制,不仅可以指定超时,还可以指定User-Agent和Referer值,这对许多网站来说都是至关重要的。

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)。

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);
    }

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

import java.io.*;
import java.net.*;

public class filedown {
    public static void download(String address, String localFileName) {
        OutputStream out = null;
        URLConnection conn = null;
        InputStream in = null;

        try {
            URL url = new URL(address);
            out = new BufferedOutputStream(new FileOutputStream(localFileName));
            conn = url.openConnection();
            in = conn.getInputStream();
            byte[] buffer = new byte[1024];

            int numRead;
            long numWritten = 0;

            while ((numRead = in.read(buffer)) != -1) {
                out.write(buffer, 0, numRead);
                numWritten += numRead;
            }

            System.out.println(localFileName + "\t" + numWritten);
        } 
        catch (Exception exception) { 
            exception.printStackTrace();
        } 
        finally {
            try {
                if (in != null) {
                    in.close();
                }
                if (out != null) {
                    out.close();
                }
            } 
            catch (IOException ioe) {
            }
        }
    }

    public static void download(String address) {
        int lastSlashIndex = address.lastIndexOf('/');
        if (lastSlashIndex >= 0 &&
        lastSlashIndex < address.length() - 1) {
            download(address, (new URL(address)).getFile());
        } 
        else {
            System.err.println("Could not figure out local file name for "+address);
        }
    }

    public static void main(String[] args) {
        for (int i = 0; i < args.length; i++) {
            download(args[i]);
        }
    }
}