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
public void saveUrl(final String filename, final String urlString)
throws MalformedURLException, IOException {
BufferedInputStream in = null;
FileOutputStream fout = null;
try {
in = new BufferedInputStream(new URL(urlString).openStream());
fout = new FileOutputStream(filename);
final byte data[] = new byte[1024];
int count;
while ((count = in.read(data, 0, 1024)) != -1) {
fout.write(data, 0, count);
}
} finally {
if (in != null) {
in.close();
}
if (fout != null) {
fout.close();
}
}
}
您将需要处理异常,可能是该方法的外部异常。
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);
}
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);
}
}
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
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);
}
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