下面是一个不起作用的代码示例:
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>
。
所以我补充说:
#[serde(deserialize_with = "Option::deserialize")]
pub info: Option<T>,
现在代码可以编译,但是尽管使用了 Option 类型,但缺少的“info”对象some_model_2
会导致错误:
thread 'main' panicked at 'called 'Result::unwrap()' on an 'Err' value: Error("missing field 'info'", line: 0, column: 0)', src\main.rs:34:6
我的最后一个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);
}
这种变化没有帮助,同样的恐慌错误仍然存在。该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。
预先感谢您的帮助!