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

第一种方法采用新通道

ReadableByteChannel aq = Channels.newChannel(new url("https//asd/abc.txt").openStream());
FileOutputStream fileOS = new FileOutputStream("C:Users/local/abc.txt")
FileChannel writech = fileOS.getChannel();

第二种方法使用FileUtils

FileUtils.copyURLToFile(new url("https//asd/abc.txt", new local file on system("C":/Users/system/abc.txt"));

第三种使用方法

InputStream xy = new ("https//asd/abc.txt").openStream();

这就是我们如何通过使用基本的Java代码和其他第三方库来下载文件。这些只是作为快速参考。请用谷歌以上关键词获取详细信息及其他选项。

其他回答

下载一个文件需要你阅读它。无论哪种方式,您都必须以某种方式查看该文件。而不是逐行,你可以从流中逐字节读取:

BufferedInputStream in = new BufferedInputStream(new URL("http://www.website.com/information.asp").openStream())
byte data[] = new byte[1024];
int count;
while((count = in.read(data, 0, 1024)) != -1)
{
    out.write(data, 0, count);
}

如果你使用代理,你可以在Java程序中设置代理,如下所示:

Properties systemSettings = System.getProperties();
systemSettings.put("proxySet", "true");
systemSettings.put("https.proxyHost", "HTTPS proxy of your org");
systemSettings.put("https.proxyPort", "8080");

如果您没有使用代理,请不要在代码中包含上述代码行。完整的工作代码下载文件时,你是一个代理。

public static void main(String[] args) throws IOException {
    String url = "https://raw.githubusercontent.com/bpjoshi/fxservice/master/src/test/java/com/bpjoshi/fxservice/api/TradeControllerTest.java";
    OutputStream outStream = null;
    URLConnection connection = null;
    InputStream is = null;
    File targetFile = null;
    URL server = null;

    // Setting up proxies
    Properties systemSettings = System.getProperties();
        systemSettings.put("proxySet", "true");
        systemSettings.put("https.proxyHost", "HTTPS proxy of my organisation");
        systemSettings.put("https.proxyPort", "8080");
        // The same way we could also set proxy for HTTP
        System.setProperty("java.net.useSystemProxies", "true");
        // Code to fetch file
    try {
        server = new URL(url);
        connection = server.openConnection();
        is = connection.getInputStream();
        byte[] buffer = new byte[is.available()];
        is.read(buffer);

        targetFile = new File("src/main/resources/targetFile.java");
        outStream = new FileOutputStream(targetFile);
        outStream.write(buffer);
    } catch (MalformedURLException e) {
        System.out.println("THE URL IS NOT CORRECT ");
        e.printStackTrace();
    } catch (IOException e) {
        System.out.println("I/O exception");
        e.printStackTrace();
    }
    finally{
        if(outStream != null)
            outStream.close();
    }
}

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();
}

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.

这是另一个基于Brian Risk的答案的Java 7变体,使用了try-with语句:

public static void downloadFileFromURL(String urlString, File destination) throws Throwable {

    URL website = new URL(urlString);
    try(
        ReadableByteChannel rbc = Channels.newChannel(website.openStream());
        FileOutputStream fos = new FileOutputStream(destination);
       ) {

        fos.getChannel().transferFrom(rbc, 0, Long.MAX_VALUE);
    }
}