gpt4 book ai didi

java - 如何使用 InputStream 从 ZIP 中读取文件?

转载 作者:IT老高 更新时间:2023-10-28 20:45:01 26 4
gpt4 key购买 nike

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

ZipFile zipFile = new ZipFile("location");

但正如我所说,我的本地计算机上没有 ZIP 文件,我不想下载它。 InputStream 是否足以读取?

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();
}


}
}

最佳答案

下面是一个关于如何提取 ZIP 文件的简单示例,您需要检查文件是否为目录。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缓冲区,然后再写入输出流。

// 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();
}
}

代码摘自以下网站:

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

关于java - 如何使用 InputStream 从 ZIP 中读取文件?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/23869228/

26 4 0
Copyright 2021 - 2024 cfsdn All Rights Reserved 蜀ICP备2022000587号
广告合作:1813099741@qq.com 6ren.com