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

这个答案几乎和选中的答案完全一样,但是有两个增强:它是一个方法,它关闭了FileOutputStream对象:

    public static void downloadFileFromURL(String urlString, File destination) {
        try {
            URL website = new URL(urlString);
            ReadableByteChannel rbc;
            rbc = Channels.newChannel(website.openStream());
            FileOutputStream fos = new FileOutputStream(destination);
            fos.getChannel().transferFrom(rbc, 0, Long.MAX_VALUE);
            fos.close();
            rbc.close();
        } catch (IOException e) {
            e.printStackTrace();
        }
    }

其他回答

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);
    }

简单使用有一个问题:

org.apache.commons.io.FileUtils.copyURLToFile(URL, File)

如果你需要下载和保存非常大的文件,或者在一般情况下,如果你需要自动重试以防连接断开。

在这种情况下,我建议使用Apache HttpClient以及org.apache.commons.io.FileUtils。例如:

GetMethod method = new GetMethod(resource_url);
try {
    int statusCode = client.executeMethod(method);
    if (statusCode != HttpStatus.SC_OK) {
        logger.error("Get method failed: " + method.getStatusLine());
    }
    org.apache.commons.io.FileUtils.copyInputStreamToFile(
        method.getResponseBodyAsStream(), new File(resource_file));
    } catch (HttpException e) {
        e.printStackTrace();
    } catch (IOException e) {
        e.printStackTrace();
    } finally {
    method.releaseConnection();
}
public class DownloadManager {

    static String urls = "[WEBSITE NAME]";

    public static void main(String[] args) throws IOException{
        URL url = verify(urls);
        HttpURLConnection connection = (HttpURLConnection) url.openConnection();
        InputStream in = null;
        String filename = url.getFile();
        filename = filename.substring(filename.lastIndexOf('/') + 1);
        FileOutputStream out = new FileOutputStream("C:\\Java2_programiranje/Network/DownloadTest1/Project/Output" + File.separator + filename);
        in = connection.getInputStream();
        int read = -1;
        byte[] buffer = new byte[4096];
        while((read = in.read(buffer)) != -1){
            out.write(buffer, 0, read);
            System.out.println("[SYSTEM/INFO]: Downloading file...");
        }
        in.close();
        out.close();
        System.out.println("[SYSTEM/INFO]: File Downloaded!");
    }
    private static URL verify(String url){
        if(!url.toLowerCase().startsWith("http://")) {
            return null;
        }
        URL verifyUrl = null;

        try{
            verifyUrl = new URL(url);
        }catch(Exception e){
            e.printStackTrace();
        }
        return verifyUrl;
    }
}

这个答案几乎和选中的答案完全一样,但是有两个增强:它是一个方法,它关闭了FileOutputStream对象:

    public static void downloadFileFromURL(String urlString, File destination) {
        try {
            URL website = new URL(urlString);
            ReadableByteChannel rbc;
            rbc = Channels.newChannel(website.openStream());
            FileOutputStream fos = new FileOutputStream(destination);
            fos.getChannel().transferFrom(rbc, 0, Long.MAX_VALUE);
            fos.close();
            rbc.close();
        } catch (IOException e) {
            e.printStackTrace();
        }
    }

你可以在一行中使用netloader for Java:

new NetFile(new File("my/zips/1.zip"), "https://example.com/example.zip", -1).load(); // Returns true if succeed, otherwise false.