实现 DeserializeOwned 的通用类型

Question

下面是一个无法运行的代码示例：

use serde_json::json;

#[derive(Serialize, Deserialize, Clone, Debug)]
pub struct Model<T>
where
    T: DeserializeOwned,
{
    pub id: i32,
    pub info: Option<T>,
}

fn main() {
    #[derive(Serialize, Deserialize, Debug, Clone)]
    struct Info {
        name: String,
    }

    let some_model_1: Model<Info> = serde_json::from_value(json!({
        "id": 43,
        "info": {
            "name": "some_model_name"
        }
    }))
    .unwrap();

    println!("some_model_1: {:#?}", some_model_1);

    let some_model_2: Model<Info> = serde_json::from_value(json!({
        "id": 43
    }))
    .unwrap();

    println!("some_model_2: {:#?}", some_model_2);
}

错误如下：cannot satisfy 'T: Deserialize<'de>.

所以我添加了：

#[serde(deserialize_with = "Option::deserialize")]
pub info: Option<T>,

现在代码可以编译，但是 some_model_2 缺少的“信息”对象导致错误，尽管使用了选项类型：

thread 'main' panicked at 'called 'Result::unwrap()' on an 'Err' value: Error("missing field 'info'", line: 0, column: 0)', src\main.rs:34:6

我最后的解决方案是使用自定义函数对 info 字段进行反序列化：

use serde::{de::DeserializeOwned, Deserialize, Deserializer, Serialize};
use serde_json::{json, Value};

#[derive(Serialize, Deserialize, Clone, Debug)]
pub struct Model<T>
where
    T: DeserializeOwned,
{
    pub id: i32,

    #[serde(deserialize_with = "ok_or_none")]
    pub info: Option<T>,
}

fn ok_or_none<'de, D, T>(deserializer: D) -> Result<Option<T>, D::Error>
where
    D: Deserializer<'de>,
    T: Deserialize<'de>,
{
    let v = Value::deserialize(deserializer)?;
    Ok(T::deserialize(v).ok())
}

fn main() {
    #[derive(Serialize, Deserialize, Debug, Clone)]
    struct Info {
        name: String,
    }

    let some_model_1: Model<Info> = serde_json::from_value(json!({
        "id": 43,
        "info": {
            "name": "some_model_name"
        }
    }))
    .unwrap();

    println!("some_model_1: {:#?}", some_model_1);

    let some_model_2: Model<Info> = serde_json::from_value(json!({
        "id": 43
    }))
    .unwrap();

    println!("some_model_2: {:#?}", some_model_2);
}

此更改无济于事，同样的恐慌错误仍然存在。 ok_or_none 函数甚至没有被调用。

以下是我的依赖项：

serde = { version = "1.0.94", features = ["derive"]  }
serde_json = "1.0.40"

我不知道我还能做些什么来完成这项工作。

提前感谢您的帮助！

Answer 1

你不需要使用 DeserializeOwned，一个简单的 T 就足够了，当从 serde 派生时它会检查你的属性是否可以是 serialized/deserialized:

use serde::{Deserialize, Serialize};
use serde_json::json;

#[derive(Serialize, Deserialize, Clone, Debug)]
pub struct Model<T> {
    pub id: i32,
    pub info: Option<T>,
}

fn main() {
    #[derive(Serialize, Deserialize, Debug, Clone)]
    struct Info {
        name: String,
    }

    let some_model_1: Model<Info> = serde_json::from_value(json!({
        "id": 43,
        "info": {
            "name": "some_model_name"
        }
    }))
    .unwrap();

    println!("some_model_1: {:#?}", some_model_1);

    let some_model_2: Model<Info> = serde_json::from_value(json!({
        "id": 43
    }))
    .unwrap();

    println!("some_model_2: {:#?}", some_model_2);
}

Playground

实现 DeserializeOwned 的通用类型

Generic type that implements DeserializeOwned

generics

json

rust

serde

serde-json