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.
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