Question

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

在官方的Actix文档中有一个示例：

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()
}

这将是最小的工作实现，到目前为止效果很好。但是您可以看到filepathstring是一个自定义字符串，该字符串将服务器上的文件重命名为upload.png（let file_path_string = "upload.png"）

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已上传文件的文件名？

Answer 1

NK所建议的content_disposition（）方法可能就是您想要的。因此，您可以替换：

let file_path_string = "upload.png";

类似：

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.")))
}

使用actix-web-framework在Rust中的网络服务器上上传文件时，使用原始文件名保存文件

1 个答案: