3

我正在使用 actix-web 框架在 Rust 中创建一个网络服务器。目前我正在处理 Fileupload，为此我正在使用 actix-multipart。

在官方 Actix-Documentation 中有一个例子：

use std::cell::Cell;
use std::fs;
use std::io::Write;

use actix_multipart::{Field, Multipart, MultipartError};
use actix_web::{error, middleware, web, App, Error, HttpResponse, HttpServer};
use futures::future::{err, Either};
use futures::{Future, Stream};

pub fn save_file(field: Field) -> impl Future<Item = i64, Error = Error> {
    let file_path_string = "upload.png";
    let file = match fs::File::create(file_path_string) {
        Ok(file) => file,
        Err(e) => return Either::A(err(error::ErrorInternalServerError(e))),
    };
    Either::B(
        field
            .fold((file, 0i64), move |(mut file, mut acc), bytes| {
                // fs operations are blocking, we have to execute writes
                // on threadpool
                web::block(move || {
                    file.write_all(bytes.as_ref()).map_err(|e| {
                        println!("file.write_all failed: {:?}", e);
                        MultipartError::Payload(error::PayloadError::Io(e))
                    })?;
                    acc += bytes.len() as i64;
                    Ok((file, acc))
                })
                .map_err(|e: error::BlockingError<MultipartError>| {
                    match e {
                        error::BlockingError::Error(e) => e,
                        error::BlockingError::Canceled => MultipartError::Incomplete,
                    }
                })
            })
            .map(|(_, acc)| acc)
            .map_err(|e| {
                println!("save_file failed, {:?}", e);
                error::ErrorInternalServerError(e)
            }),
    )
}

pub fn upload(
    multipart: Multipart,
    counter: web::Data<Cell<usize>>,
) -> impl Future<Item = HttpResponse, Error = Error> {
    counter.set(counter.get() + 1);
    println!("{:?}", counter.get());

    multipart
        .map_err(error::ErrorInternalServerError)
        .map(|field| save_file(field).into_stream())
        .flatten()
        .collect()
        .map(|sizes| HttpResponse::Ok().json(sizes))
        .map_err(|e| {
            println!("failed: {}", e);
            e
        })
}

fn index() -> HttpResponse {
    let html = r#"<html>
        <head><title>Upload Test</title></head>
        <body>
            <form target="/" method="post" enctype="multipart/form-data">
                <input type="file" name="file"/>
                <input type="submit" value="Submit"></button>
            </form>
        </body>
    </html>"#;

    HttpResponse::Ok().body(html)
}

fn main() -> std::io::Result<()> {

    HttpServer::new(|| {
        App::new()
            .data(Cell::new(0usize))
            .wrap(middleware::Logger::default())
            .service(
                web::resource("/")
                    .route(web::get().to(index))
                    .route(web::post().to_async(upload)),
            )
    })
    .bind("127.0.0.1:8080")?
    .run()
}

这将是它的最小工作实现，并且到目前为止运行良好。但正如您所见，filepathstring 是一个自定义字符串，它将服务器上的文件重命名为 upload.png ( let file_path_string = "upload.png")

那么有没有一种简单的方法来检索原始文件名并将其用作服务器上上传文件的文件名？

4

1 回答 1

2

NK 建议的 content_disposition() 方法可能是您在这里所追求的。所以你也许可以替换：

let file_path_string = "upload.png";

有类似的东西：

let file_path_string = match field.content_disposition().unwrap().get_filename() {
    Some(filename) => filename.replace(' ', "_").to_string(),
    None => return Either::A(err(error::ErrorInternalServerError("Couldn't read the filename.")))
}
于 2019-05-18T18:40:59.623 回答