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。

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。

在underscore-java库中有一个方法U.fetch(url)。

文件pom.xml:

<dependency>
  <groupId>com.github.javadev</groupId>
  <artifactId>underscore</artifactId>
  <version>1.84</version>
</dependency>

代码示例:

import com.github.underscore.U;
import java.io.IOException;
import java.nio.file.Files;
import java.nio.file.Paths;

public class Download {
    public static void main(String[] args) throws IOException {
        Files.write(Paths.get("data.bin"),
            U.fetch("https://stackoverflow.com/questions"
                + "/921262/how-to-download-and-save-a-file-from-internet-using-java").blob());
    }
}

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)。

在java.net.http.HttpClient上使用授权的解决方案:

HttpClient client = HttpClient.newHttpClient();

HttpRequest request = HttpRequest.newBuilder()
        .GET()
        .header("Accept", "application/json")
        // .header("Authorization", "Basic ci5raG9kemhhZXY6NDdiYdfjlmNUM=") if you need
        .uri(URI.create("https://jira.google.ru/secure/attachment/234096/screenshot-1.png"))
        .build();

HttpResponse<InputStream> response = client.send(request, HttpResponse.BodyHandlers.ofInputStream());

try (InputStream in = response.body()) {
    Files.copy(in, Paths.get(target + "screenshot-1.png"), StandardCopyOption.REPLACE_EXISTING);
}

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);
    }
}