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
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]);
}
}
}
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。
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);
}
}
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]);
}
}
}
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
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