如何从zip文件中读取特定文件

Question

我完全卡住了从zip文件的可变路径结构中读取文件而不解压缩它。

我的文件位于这里:

/info/[random-string]/info.json

其中[random-string]是info文件夹中唯一的文件。就像读取'info'文件夹读取第一个文件夹读取'info.json'。

任何想法如何做到这一点与这些库之一(zip或rc_zip)?

let file_path = file.to_str().unwrap();
let file = File::open(file_path).unwrap();
let reader = BufReader::new(file);
let mut archive = zip::ZipArchive::new(reader).unwrap();
let info_folder = archive.by_name("info").unwrap();
// how to list files of info_folder

Answer 1

给你:

use std::error::Error;
use std::ffi::OsStr;
use std::fs::File;
use std::path::Path;
use zip::ZipArchive; // zip 0.5.13
fn main() -> Result<(), Box<dyn Error>> {
let archive = File::open("./info.zip")?;
let mut archive = ZipArchive::new(archive)?;
// iterate over all files, because you don't know the exact name
for idx in 0..archive.len() {
let entry = archive.by_index(idx)?;
let name = entry.enclosed_name();

if let Some(name) = name {
// process only entries which are named  info.json
if name.file_name() == Some(OsStr::new("info.json")) {
// the ancestors() iterator lets you walk up the path segments
let mut ancestors = name.ancestors();

// skip self - the first entry is always the full path
ancestors.next(); 
// skip the random string
ancestors.next(); 
let expect_info = ancestors.next(); 
// the reminder must be only 'info/' otherwise this is the wrong entry
if expect_info == Some(Path::new("info/")) {
// do something with the file
println!("Found!!!");
break;
}
}
}
}
Ok(())
}