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
这可以读取互联网上的文件,并将其写入文件。
import java.net.URL;
import java.io.FileOutputStream;
import java.io.File;
public class Download {
public static void main(String[] args) throws Exception {
URL url = new URL("https://www.google.com/images/branding/googlelogo/1x/googlelogo_color_272x92dp.png"); // Input URL
FileOutputStream out = new FileOutputStream(new File("out.png")); // Output file
out.write(url.openStream().readAllBytes());
out.close();
}
}
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。
import java.io.*;
import java.net.*;
public class filedown {
public static void download(String address, String localFileName) {
OutputStream out = null;
URLConnection conn = null;
InputStream in = null;
try {
URL url = new URL(address);
out = new BufferedOutputStream(new FileOutputStream(localFileName));
conn = url.openConnection();
in = conn.getInputStream();
byte[] buffer = new byte[1024];
int numRead;
long numWritten = 0;
while ((numRead = in.read(buffer)) != -1) {
out.write(buffer, 0, numRead);
numWritten += numRead;
}
System.out.println(localFileName + "\t" + numWritten);
}
catch (Exception exception) {
exception.printStackTrace();
}
finally {
try {
if (in != null) {
in.close();
}
if (out != null) {
out.close();
}
}
catch (IOException ioe) {
}
}
}
public static void download(String address) {
int lastSlashIndex = address.lastIndexOf('/');
if (lastSlashIndex >= 0 &&
lastSlashIndex < address.length() - 1) {
download(address, (new URL(address)).getFile());
}
else {
System.err.println("Could not figure out local file name for "+address);
}
}
public static void main(String[] args) {
for (int i = 0; i < args.length; i++) {
download(args[i]);
}
}
}
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.
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)。
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();
}
}
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