我有一个rusoto_core::ByteStream
实现了futures' Stream
trait:
let chunks = vec![b"1234".to_vec(), b"5678".to_vec()];
let stream = ByteStream::new(stream::iter_ok(chunks));
我想将其传递给actix_web's HttpResponseBuilder::streaming
方法。
use actix_web::dev::HttpResponseBuilder; // 0.7.18
use rusoto_core::ByteStream; // 0.36.0
fn example(stream: ByteStream, builder: HttpResponseBuilder) {
builder.streaming(stream);
}
当我尝试执行此操作时,出现以下错误:
error[E0271]: type mismatch resolving `<rusoto_core::stream::ByteStream as futures::stream::Stream>::Item == bytes::bytes::Bytes`
--> src/main.rs:5:13
|
5 | builder.streaming(stream);
| ^^^^^^^^^ expected struct `std::vec::Vec`, found struct `bytes::bytes::Bytes`
|
= note: expected type `std::vec::Vec<u8>`
found type `bytes::bytes::Bytes`
我相信原因是streaming()
期望有S: Stream<Item = Bytes, Error>
(即Item = Bytes
),但是我的ByteStream
有Item = Vec<u8>
。我该如何解决?
我认为解决方案是以某种方式flatmap
我的ByteStream
,但我找不到这种流方法。
下面是一个如何使用streaming()
的示例:
let text = "123";
let (tx, rx_body) = mpsc::unbounded();
let _ = tx.unbounded_send(Bytes::from(text.as_bytes()));
HttpResponse::Ok()
.streaming(rx_body.map_err(|e| error::ErrorBadRequest("bad request")))
答案 0 :(得分:4)
如何
flatmap
在Rust中进行流式播放?
平面图将迭代器的迭代器转换为单个迭代器(或用流代替迭代器)。
Futures 0.1没有直接的平面地图,但是确实有Stream::flatten
,可以在Stream::map
之后使用。
use futures::{stream, Stream}; // 0.1.25
fn into_many(i: i32) -> impl Stream<Item = i32, Error = ()> {
stream::iter_ok(0..i)
}
fn nested() -> impl Stream<Item = i32, Error = ()> {
let stream_of_number = into_many(5);
let stream_of_stream_of_number = stream_of_number.map(into_many);
let flat_stream_of_number = stream_of_stream_of_number.flatten();
// Returns: 0, 0, 1, 0, 1, 2, 0, 1, 2, 3
flat_stream_of_number
}
但是,这不能解决您的问题。
streaming()
期望有S: Stream<Item = Bytes, Error>
(即Item = Bytes
),但是我的ByteStream
有Item = Vec<u8>
是的,这就是问题所在。通过Bytes::from
使用Stream::map
将流Item
从一种类型转换为另一种类型:
use bytes::Bytes; // 0.4.11
use futures::Stream; // 0.1.25
fn example(stream: ByteStream, mut builder: HttpResponseBuilder) {
builder.streaming(stream.map(Bytes::from));
}