Sto cercando di leggere direttamente un file zip da un URL remoto Ho cercato in questo modoCome leggere un file zip da un URL remoto senza estrarlo
import java.io.BufferedInputStream;
import java.io.BufferedReader;
import java.io.File;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.net.URL;
import java.util.zip.ZipEntry;
import java.util.zip.ZipFile;
import java.util.zip.ZipInputStream;
public class Utils {
public static void main(String args[]) throws Exception {
String ftpUrl = "http://wwwccc.zip";
URL url = new URL(ftpUrl);
unpackArchive(url);
}
public static void unpackArchive(URL url) throws IOException {
String ftpUrl = "http://www.vvvv.xip";
File zipFile = new File(url.toString());
ZipFile zip = new ZipFile(zipFile);
InputStream in = new BufferedInputStream(url.openStream(), 1024);
ZipInputStream zis = new ZipInputStream(in);
ZipEntry entry;
while ((entry = zis.getNextEntry()) != null) {
System.out.println("entry: " + entry.getName() + ", "
+ entry.getSize());
BufferedReader bufferedeReader = new BufferedReader(
new InputStreamReader(zip.getInputStream(entry)));
String line = bufferedeReader.readLine();
while (line != null) {
System.out.println(line);
line = bufferedeReader.readLine();
}
bufferedeReader.close();
}
}
}
sto ottenendo eccezione come
Exception in thread "main" java.io.FileNotFoundException: http:\www.nseindia.com\content\historical\EQUITIES\2015\NOV\cm03NOV2015bhav.csv.zip (The filename, directory name, or volume label syntax is incorrect)
at java.util.zip.ZipFile.open(Native Method)
at java.util.zip.ZipFile.<init>(Unknown Source)
at java.util.zip.ZipFile.<init>(Unknown Source)
at java.util.zip.ZipFile.<init>(Unknown Source)
at Utils.unpackArchive(Utils.java:30)
at Utils.main(Utils.java:19)
Dove l'URL del file zip funziona correttamente quando si esegue da un browser.
Grazie mille, mi hai salvato. – Pawan