我有以下格式的 JSON 对象:
{
"name": "foo",
"value": 1234,
"upper_bound": 5000,
"lower_bound": 1000
}
我想使用 serde 来处理这些对象,其结构如下
struct MyObject {
name: String,
value: i32,
bound: Range<i32>,
}
无需任何修改,序列化这些结构之一就会产生
{
"name": "foo",
"value": 1234,
"bound": {
"start": 1000,
"end": 5000
}
}
我可以应用#[serde(flatten)]
来更接近,产生
{
"name": "foo",
"value": 1234,
"start": 1000,
"end": 5000
}
但是添加 #[serde(rename...)]
似乎没有改变任何内容,无论我尝试为重命名提供什么样的参数。是否可以展平范围并重命名参数?
最佳答案
您可以使用 serde 属性 with
并仅使用中间结构让真正的实现进行 serde:
use core::ops::Range;
use serde::{Deserialize, Serialize};
use serde_json::Error;
#[derive(Debug, Default, PartialEq, Eq, Serialize, Deserialize)]
struct Foo {
name: String,
value: i32,
#[serde(with = "range_aux", flatten)]
bound: Range<i32>,
}
mod range_aux {
use core::ops::Range;
use serde::{Deserialize, Deserializer, Serialize, Serializer};
#[derive(Serialize, Deserialize)]
struct RangeAux {
upper_bound: i32,
lower_bound: i32,
}
pub fn serialize<S>(range: &Range<i32>, ser: S) -> Result<S::Ok, S::Error>
where
S: Serializer,
{
RangeAux::serialize(
&RangeAux {
upper_bound: range.end,
lower_bound: range.start,
},
ser,
)
}
pub fn deserialize<'de, D>(d: D) -> Result<Range<i32>, D::Error>
where
D: Deserializer<'de>,
{
let range_aux: RangeAux = RangeAux::deserialize(d)?;
Ok(Range {
start: range_aux.lower_bound,
end: range_aux.upper_bound,
})
}
}
fn main() -> Result<(), Error> {
let data = r#"{"name":"foo","value":1234,"upper_bound":5000,"lower_bound":1000}"#;
let foo: Foo = serde_json::from_str(data)?;
assert_eq!(
foo,
Foo {
name: "foo".to_string(),
value: 1234,
bound: 1000..5000
}
);
let output = serde_json::to_string(&foo)?;
assert_eq!(data, output);
Ok(())
}
非常接近remote
模式但这不适用于通用请参阅 serde#1844
.
可能的通用版本:
use core::ops::Range;
use serde::{Deserialize, Serialize};
use serde_json::Error;
#[derive(Debug, Default, PartialEq, Eq, Serialize, Deserialize)]
struct Foo {
name: String,
value: i32,
#[serde(with = "range_aux", flatten)]
bound: Range<i32>,
}
mod range_aux {
use core::ops::Range;
use serde::{Deserialize, Deserializer, Serialize, Serializer};
pub fn serialize<S, Idx: Serialize>(range: &Range<Idx>, ser: S) -> Result<S::Ok, S::Error>
where
S: Serializer,
{
// could require Idx to be Copy or Clone instead of borrowing Idx
#[derive(Serialize)]
struct RangeAux<'a, Idx> {
upper_bound: &'a Idx,
lower_bound: &'a Idx,
}
RangeAux::serialize(
&RangeAux {
upper_bound: &range.end,
lower_bound: &range.start,
},
ser,
)
}
pub fn deserialize<'de, D, Idx: Deserialize<'de>>(d: D) -> Result<Range<Idx>, D::Error>
where
D: Deserializer<'de>,
{
#[derive(Deserialize)]
struct RangeAux<Idx> {
upper_bound: Idx,
lower_bound: Idx,
}
let range_aux: RangeAux<Idx> = RangeAux::deserialize(d)?;
Ok(Range {
start: range_aux.lower_bound,
end: range_aux.upper_bound,
})
}
}
fn main() -> Result<(), Error> {
let data = r#"{"name":"foo","value":1234,"upper_bound":5000,"lower_bound":1000}"#;
let foo: Foo = serde_json::from_str(data)?;
assert_eq!(
foo,
Foo {
name: "foo".to_string(),
value: 1234,
bound: 1000..5000
}
);
let output = serde_json::to_string(&foo)?;
assert_eq!(data, output);
Ok(())
}
关于rust - 如何使用 serde 重命名 `start` 和 `end` 范围值?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/72483890/