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
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);
}
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。
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);
}
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.
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;
}
}
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