Question

我正在尝试直接从远程URL读取zip文件我试过这种方式

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();
        }
    }
}

我的异常是

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)

从浏览器运行时，zip文件的URL工作正常。

Answer 1

HttpURLConnection类不适用于远程文件。它仅支持本地文件系统上可用的文件。要在远程文件上打开流，您可以使用HttpURLConnection。

在String url= "http://www.nseindia.com/content/historical/EQUITIES/2015/NOV/cm03NOV2015bhav.csv.zip"; InputStream is = new URL(url).openConnection().getInputStream();实例上调用getInputStream()以获取可以进一步处理的输入流。

示例：

{{1}}

Answer 2

以上都不适合我。

做了什么 像魅力一样工作，是这样的：

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

Answer 3

使用

行

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

您正在尝试创建一个名为URL的文件，其中包含不允许使用的字符。

该文件的名称应该更简单，如

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

编译器告诉你，以及：

（文件名，目录名或卷标语法不正确）

我确信这就是你收到错误的原因。但我不确定其余的代码。

Answer 4

读取存储在远程位置的Zip文件。只需在下面的代码库中更改您的远程位置的zip文件详细信息即可。

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();
  }

}

Answer 5

你不能像这样从url创建文件试试这个：

网址ftpUrl =新网址（“http://www.nseindia.com/content/historical/EQUITIES＆gt; /2015/NOV/cm03NOV2015bhav.csv.zip”）;

文件zipFile =新文件（“本地驱动器上的某个位置”）;

FileUtils.copyURLToFile（ftpUrl，zipFile）;

ZipFile zip = new ZipFile（zipFile）;

如何从远程URL读取zip文件而不提取它

5 个答案: