CODEWITHSUNDEEP
java
Pawan
Pawan

Reputation: 32321

How to read a zip file from a remote URL without extracting it

I am trying to directly read a zip file from a Remote URL I have tried this way 

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();
        }
    }
}

I am getting Exception as

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)

Where as the URL of zip file is working fine when running from a browser .

Upvotes: 9

Views: 13371

Answers (5)

pintu
pintu

Reputation: 331

Reading Zip file stored on remote Location.Just need to change your remote location zip file Details in below code base.

import java.io.*;
import java.net.URL;
import java.util.zip.ZipEntry;
import java.util.zip.ZipInputStream;

  public class ReadZipFileFromRemote{
        public static void main(String args[]){
           String url="https://test-po.s3.ap-south-1.amazonaws.com/dev/coding/76/55/14/1/1587736321256.zip";
           String content=readZipFileFromRemote(url);
           System.out.println(content);  
        }  

        public String readZipFileFromRemote(String remoteFileUrl) {        

           StringBuilder sb = new StringBuilder();
         try {
             URL url = new URL(remoteFileUrl);
             InputStream in = new BufferedInputStream(url.openStream(), 1024);
             ZipInputStream stream = new ZipInputStream(in);
             byte[] buffer = new byte[1024];
             ZipEntry entry;
             while ((entry = stream.getNextEntry()) != null) {
                  int read;
                 while ((read = stream.read(buffer, 0, 1024)) >= 0) {
                      sb.append(new String(buffer, 0, read));
                }
            }
     } catch (Exception e) {
        e.printStackTrace();
     }
    return sb.toString();
  }

}

Upvotes: 3

david a.
david a.

Reputation: 5291

File class is not designed to work with remote files. It only supports files that are available on a local file system. To open a stream on a remote file, you can use HttpURLConnection.

Call getInputStream() on an HttpURLConnection instance to get an input stream that you can process further.

Example:

String url= "http://www.nseindia.com/content/historical/EQUITIES/2015/NOV/cm03NOV2015bhav.csv.zip";
InputStream is = new URL(url).openConnection().getInputStream();

Upvotes: 3

Mike Nakis
Mike Nakis

Reputation: 61986

None of the above has worked for me.

What did work like a charm, is this:

InputStream inputStream = new URL( urlString ).openStream();

Upvotes: 2

Partha Dalai
Partha Dalai

Reputation: 9

you cant create file from url like that try this :

URL ftpUrl = new URL("http://www.nseindia.com/content/historical/EQUITIES>/2015/NOV/cm03NOV2015bhav.csv.zip");

File zipFile = new File("some location on your local drive");

FileUtils.copyURLToFile(ftpUrl, zipFile);

ZipFile zip = new ZipFile(zipFile);

Upvotes: -3

MiKE
MiKE

Reputation: 524

With the line 

File zipFile = new File(url.toString());

you are trying to create a file named like the URL which contains characters that are not allowed.

The file should be named simpler like

File zipFile = new File("zipfile.csv.zip");

The compiler is telling you that aswell:

(The filename, directory name, or volume label syntax is incorrect)

I am sure this is why you are getting the error. But i'm not certain about the rest of the code.

Upvotes: 1

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