我想chrono::NaiveDate
用自定义函数对 a 进行序列化和反序列化,但 Serde 书并未涵盖此功能,代码文档也无济于事。
#[macro_use]
extern crate serde_derive;
extern crate serde;
extern crate serde_json;
extern crate chrono;
use chrono::NaiveDate;
mod date_serde {
use chrono::NaiveDate;
use serde::{self, Deserialize, Serializer, Deserializer};
pub fn serialize<S>(date: &Option<NaiveDate>, s: S) -> Result<S::Ok, S::Error>
where S: Serializer {
if let Some(ref d) = *date {
return s.serialize_str(&d.format("%Y-%m-%d").to_string())
}
s.serialize_none()
}
pub fn deserialize<'de, D>(deserializer: D)
-> Result<Option<NaiveDate>, D::Error>
where D: Deserializer<'de> {
let s: Option<String> = Option::deserialize(deserializer)?;
if let Some(s) = s {
return Ok(Some(NaiveDate::parse_from_str(&s, "%Y-%m-%d").map_err(serde::de::Error::custom)?))
}
Ok(None)
}
}
#[derive(Debug, Serialize, Deserialize)]
struct Test {
pub i: u64,
#[serde(with = "date_serde")]
pub date: Option<NaiveDate>,
}
fn main() {
let mut test: Test = serde_json::from_str(r#"{"i": 3, "date": "2015-02-03"}"#).unwrap();
assert_eq!(test.i, 3);
assert_eq!(test.date, Some(NaiveDate::from_ymd(2015, 02, 03)));
test = serde_json::from_str(r#"{"i": 5}"#).unwrap();
assert_eq!(test.i, 5);
assert_eq!(test.date, None);
}
我知道这Option<chrono::NaiveDate>
很容易被 Serde 反序列化,因为 Chrono 支持 Serde,但我正在尝试学习 Serde,所以我想自己实现它。当我运行此代码时,出现错误:
thread 'main' panicked at 'called `Result::unwrap()` on an `Err` value: ErrorImpl { code: Message("missing field `date`"), line: 1, column: 8 }', /checkout/src/libcore/result.rs:859
note: Run with `RUST_BACKTRACE=1` for a backtrace.