首页 > 解决方案 > 在 actix-web 中用作查询参数时,如何将无效的枚举变体变为无

问题描述

使用文档中提供的示例,在提供未知变体时如何使用actix_web::web::Queryresponse_typeNone

如果我有以下情况:

use actix_web::{web, App, HttpServer};
use serde::Deserialize;

#[derive(Debug, Deserialize)]
pub enum ResponseType {
    Token,
    Code,
}

#[derive(Deserialize)]
pub struct AuthRequest {
    id: u64,
    response_type: Option<ResponseType>,
}

async fn index(web::Query(info): web::Query<AuthRequest>) -> String {
    format!(
        "Authorization request for client with id={} and type={:?}!",
        info.id, info.response_type
    )
}

#[actix_web::main]
async fn main() -> std::io::Result<()> {
    HttpServer::new(|| App::new().route("/", web::get().to(index)))
        .bind(("127.0.0.1", 8080))?
        .run()
        .await
}

我访问http://localhost:8080/?id=1&response_type=foo,我得到这个 400 响应:

查询反序列化错误:未知变体foo,预期TokenCode

当我希望它只接受 Enum 的值作为有效值时,如果没有提供值或无效值,我希望它设置为None.

标签: rustserdeactix-web

解决方案


这可以通过 deserialize_with 来处理

use actix_web::{web, App, HttpServer};
use serde::Deserialize;
use serde::de::{Deserializer};

#[derive(Debug, Deserialize)]
pub enum ResponseType {
    Token,
    Code,
}

fn from_response_type<'de, D>(deserializer: D) -> Result<Option<ResponseType>, D::Error>
where
    D: Deserializer<'de>,
{
    let res: Option<ResponseType> = Deserialize::deserialize(deserializer).unwrap_or(None);
    Ok(res)
}

#[derive(Debug, Deserialize)]
pub struct AuthRequest {
    id: u64,
    #[serde(deserialize_with = "from_response_type")]
    response_type: Option<ResponseType>,
}

async fn index(web::Query(info): web::Query<AuthRequest>) -> String {
    format!(
        "Authorization request for client with id={} and type={:?}!",
        info.id, info.response_type
    )
}

#[actix_web::main]
async fn main() -> std::io::Result<()> {
    HttpServer::new(|| App::new().route("/", web::get().to(index)))
        .bind(("127.0.0.1", 8080))?
        .run()
        .await
}

任何无效值都被视为None. 关键线是

let res: Option<ResponseType> = Deserialize::deserialize(deserializer).unwrap_or(None);

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