Question

我必须使用SFTP从ZIP存档（只有一个文件，我知道它的名称）中获取文件内容。我唯一拥有的是ZIP InputStream。大多数示例显示如何使用此语句获取内容：

ZipFile zipFile = new ZipFile("location");

但正如我所说，我的本地机器上没有ZIP文件，我也不想下载它。 InputStream是否足以阅读？

UPD：我就是这样做的：

import java.util.zip.ZipInputStream;

import com.jcraft.jsch.Channel;
import com.jcraft.jsch.ChannelSftp;
import com.jcraft.jsch.JSch;
import com.jcraft.jsch.Session;

public class SFTP {


    public static void main(String[] args) {

        String SFTPHOST = "host";
        int SFTPPORT = 3232;
        String SFTPUSER = "user";
        String SFTPPASS = "mypass";
        String SFTPWORKINGDIR = "/dir/work";
        Session session = null;
        Channel channel = null;
        ChannelSftp channelSftp = null;
        try {
            JSch jsch = new JSch();
            session = jsch.getSession(SFTPUSER, SFTPHOST, SFTPPORT);
            session.setPassword(SFTPPASS);
            java.util.Properties config = new java.util.Properties();
            config.put("StrictHostKeyChecking", "no");
            session.setConfig(config);
            session.connect();
            channel = session.openChannel("sftp");
            channel.connect();
            channelSftp = (ChannelSftp) channel;
            channelSftp.cd(SFTPWORKINGDIR);
            ZipInputStream stream = new ZipInputStream(channelSftp.get("file.zip"));
            ZipEntry entry = zipStream.getNextEntry();
            System.out.println(entry.getName); //Yes, I got its name, now I need to get content
        } catch (Exception ex) {
            ex.printStackTrace();
        } finally {
            session.disconnect();
            channelSftp.disconnect();
            channel.disconnect();
        }


    }
}

Answer 1

下面是一个关于如何提取ZIP文件的简单示例，您需要检查该文件是否是目录。但这是最简单的。

您缺少的步骤是读取输入流并将内容写入写入输出流的缓冲区。

// Expands the zip file passed as argument 1, into the
// directory provided in argument 2
public static void main(String args[]) throws Exception
{
    if(args.length != 2)
    {
        System.err.println("zipreader zipfile outputdir");
        return;
    }

    // create a buffer to improve copy performance later.
    byte[] buffer = new byte[2048];

    // open the zip file stream
    InputStream theFile = new FileInputStream(args[0]);
    ZipInputStream stream = new ZipInputStream(theFile);
    String outdir = args[1];

    try
    {

        // now iterate through each item in the stream. The get next
        // entry call will return a ZipEntry for each file in the
        // stream
        ZipEntry entry;
        while((entry = stream.getNextEntry())!=null)
        {
            String s = String.format("Entry: %s len %d added %TD",
                            entry.getName(), entry.getSize(),
                            new Date(entry.getTime()));
            System.out.println(s);

            // Once we get the entry from the stream, the stream is
            // positioned read to read the raw data, and we keep
            // reading until read returns 0 or less.
            String outpath = outdir + "/" + entry.getName();
            FileOutputStream output = null;
            try
            {
                output = new FileOutputStream(outpath);
                int len = 0;
                while ((len = stream.read(buffer)) > 0)
                {
                    output.write(buffer, 0, len);
                }
            }
            finally
            {
                // we must always close the output file
                if(output!=null) output.close();
            }
        }
    }
    finally
    {
        // we must always close the zip file.
        stream.close();
    }
}

代码摘录来自以下网站：

http://www.thecoderscorner.com/team-blog/java-and-jvm/12-reading-a-zip-file-from-java-using-zipinputstream#.U4RAxYamixR

Answer 2

好吧，我已经做到了：

 zipStream = new ZipInputStream(channelSftp.get("Port_Increment_201405261400_2251.zip"));
 zipStream.getNextEntry();

 sc = new Scanner(zipStream);
 while (sc.hasNextLine()) {
     System.out.println(sc.nextLine());
 }

它帮助我阅读ZIP的内容，而无需写入另一个文件。

Answer 3

ZipInputStream本身就是InputStream，并在每次调用getNextEntry()后传递每个条目的内容。必须特别小心，不要关闭读取内容的流，因为它与ZIP流相同：

public void readZipStream(InputStream in) throws IOException {
    ZipInputStream zipIn = new ZipInputStream(in);
    ZipEntry entry;
    while ((entry = zipIn.getNextEntry()) != null) {
        System.out.println(entry.getName());
        readContents(zipIn);
        zipIn.closeEntry();
    }
}

private void readContents(InputStream contentsIn) throws IOException {
    byte contents[] = new byte[4096];
    int direct;
    while ((direct = contentsIn.read(contents, 0, contents.length)) >= 0) {
        System.out.println("Read " + direct + "bytes content.");
    }
}

将阅读内容委托给其他逻辑时，可能需要用ZipInputStream包裹FilterInputStream以仅关闭条目而不是整个流，如下所示：

public void readZipStream(InputStream in) throws IOException {
    ZipInputStream zipIn = new ZipInputStream(in);
    ZipEntry entry;
    while ((entry = zipIn.getNextEntry()) != null) {
        System.out.println(entry.getName());

        readContents(new FilterInputStream(zipIn) {
            @Override
            public void close() throws IOException {
                zipIn.closeEntry();
            }
        });
    }
}

Answer 4

OP接近了。只需要读取字节。调用getNextEntry positions the stream at the beginning of the entry data（docs）。如果这是我们想要的条目（或唯一的条目），则InputStream在正确的位置。我们所需要做的就是读取该条目的解压缩字节。

byte[] bytes = new byte[(int) entry.getSize()];
int i = 0;
while (i < bytes.length) {
    // .read doesn't always fill the buffer we give it.
    // Keep calling it until we get all the bytes for this entry.
    i += zipStream.read(bytes, i, bytes.length - i);
}

因此，如果这些字节确实是文本，那么我们可以将这些字节解码为String。我只是假设使用utf8编码。

new String(bytes, "utf8")

旁注：我个人使用apache commons-io IOUtils来减少此类较低级别的内容。 ZipInputStream.read的文档似乎暗示着读取将在当前zip条目的末尾停止。如果是这样，那么读取当前文本条目就是使用IOUtils的一行。

String text = IOUtils.toString(zipStream)

Answer 5

这里是使用BiConsumer处理zip输入流的更通用的解决方案。这几乎与haui使用的解决方案相同

private void readZip(InputStream is, BiConsumer<ZipEntry,InputStream> consumer) throws IOException {
    try (ZipInputStream zipFile = new ZipInputStream(is);) {
        ZipEntry entry;
        while((entry = zipFile.getNextEntry()) != null){
            consumer.accept(entry, new FilterInputStream(zipFile) {
                @Override
                public void close() throws IOException {
                    zipFile.closeEntry();
                }
            });
        }
    }
}

您可以通过调用

来使用它

readZip(<some inputstream>, (entry, is) -> {
    /* don't forget to close this stream after processing. */
    is.read() // ... <- to read each entry
});

Answer 6

将包含文件结构的压缩文件（zip）解压缩到给定目录中。注意;该代码在“ org.apache.commons.io.IOUtils”上使用deps，但您可以将其替换为您的自定义“读取流”代码

public static void unzipDirectory(File archiveFile, File destinationDir) throws IOException
{
  Path destPath = destinationDir.toPath();
  try (ZipInputStream zis = new ZipInputStream(new FileInputStream(archiveFile)))
  {
    ZipEntry zipEntry;
    while ((zipEntry = zis.getNextEntry()) != null)
    {
      Path resolvedPath = destPath.resolve(zipEntry.getName()).normalize();
      if (!resolvedPath.startsWith(destPath))
      {
        throw new IOException("The requested zip-entry '" + zipEntry.getName() + "' does not belong to the requested destination");
      }
      if (zipEntry.isDirectory())
      {
        Files.createDirectories(resolvedPath);
      } else
      {
        if(!Files.isDirectory(resolvedPath.getParent()))
        {
          Files.createDirectories(resolvedPath.getParent());
        }
        try (FileOutputStream outStream = new FileOutputStream(resolvedPath.toFile()))
        {
          IOUtils.copy(zis, outStream);
        }
      }
    }
  }
}

Answer 7

如果ZIP的内容包含1个文件（例如，HTTP响应的压缩内容），则可以使用Kotlin读取文本内容，如下所示：

@Throws(IOException::class)
fun InputStream.readZippedContent() = ZipInputStream(this).use { stream ->
     stream.nextEntry?.let { stream.bufferedReader().readText() } ?: String()
}

此扩展功能可解压缩Zip文件的第一个ZIP条目，并以纯文本格式读取内容。

用法：

val inputStream: InputStream = ... // your zipped InputStream
val textContent = inputStream.readZippedContent()

如何使用InputStream从ZIP读取文件？

7 个答案: