此解决方案使用RawValue
from 类型serde_json
首先获取原始输入字符串。然后Deserializer
从该字符串创建一个新的以反序列化该User
类型。
该解决方案可以通过Box<serde_json::value::RawValue>
用作中介类型与读取器一起使用,也可以通过用作中介来与从输入借用的结构一起使用&'de serde_json::value::RawValue
。您可以通过(取消)注释该borrow
字段在解决方案中对其进行测试。
use std::marker::PhantomData;
#[derive(Debug, serde::Serialize, serde::Deserialize)]
#[serde(remote = "Self")]
struct User<'a> {
#[serde(skip)]
pub raw: String,
pub id: u64,
pub login: String,
// Test for borrowing input data
// pub borrow: &'a str,
#[serde(skip)]
pub ignored: PhantomData<&'a ()>,
}
impl serde::Serialize for User<'_> {
fn serialize<S: serde::Serializer>(&self, serializer: S) -> Result<S::Ok, S::Error> {
Self::serialize(self, serializer)
}
}
impl<'a, 'de> serde::Deserialize<'de> for User<'a>
where
'de: 'a,
{
fn deserialize<D: serde::Deserializer<'de>>(deserializer: D) -> Result<Self, D::Error> {
use serde::de::Error;
// Deserializing a `&'a RawValue` would also work here
// but then you loose support for deserializing from readers
let raw: Box<serde_json::value::RawValue> = Box::deserialize(deserializer)?;
// Use this line instead if you have a struct which borrows from the input
// let raw = <&'de serde_json::value::RawValue>::deserialize(deserializer)?;
let mut raw_value_deserializer = serde_json::Deserializer::from_str(raw.get());
let mut user =
User::deserialize(&mut raw_value_deserializer).map_err(|err| D::Error::custom(err))?;
user.raw = raw.get().to_string();
Ok(user)
}
}
fn main() {
// Test serialization
let u = User {
raw: String::new(),
id: 456,
login: "USERNAME".to_string(),
// Test for borrowing input data
// borrow: "foobar",
ignored: PhantomData,
};
let json = serde_json::to_string(&u).unwrap();
println!("{}", json);
// Test deserialization
let u2: User = serde_json::from_str(&json).unwrap();
println!("{:#?}", u2);
}
在操场上测试。