#[serde(rename)]似乎是正确的 Select ，但文档中没有说明是否可能或如何做到这一点.
这个JSON对象:
{
"name" : "myobject"
"info" :
{
"counter" : "3"
"foo" : "bar"
}
}
对应的flat rust struct 应为:
#[derive(Deserialize)]
struct Object {
name: String,
#[serde(rename="info.counter")] // wrong syntax here !!
count: i32,
#[serde(rename="info::foo")] // neither this works
foo: String,
}
对于属性，没有内置的方法可以做到这一点，但是您可以为Object类型编写自己的Deserialize impl，它首先反序列化到某个中间助手表示，然后将数据重新排列到所需的 struct 中.
#[macro_use]
extern crate serde_derive;
extern crate serde;
extern crate serde_json;
use serde::{Deserialize, Deserializer};
#[derive(Debug)]
struct Object {
name: String,
count: i32,
foo: String,
}
impl<'de> Deserialize<'de> for Object {
fn deserialize<D>(deserializer: D) -> Result<Self, D::Error>
where D: Deserializer<'de>
{
#[derive(Deserialize)]
struct Outer {
name: String,
info: Inner,
}
#[derive(Deserialize)]
struct Inner {
count: i32,
foo: String,
}
let helper = Outer::deserialize(deserializer)?;
Ok(Object {
name: helper.name,
count: helper.info.count,
foo: helper.info.foo,
})
}
}
fn main() {
let j = r#"{
"name": "myobject",
"info": {
"count": 3,
"foo": "bar"
}
}"#;
println!("{:#?}", serde_json::from_str::<Object>(j).unwrap());
}
输出为:
Object {
name: "myobject",
count: 3,
foo: "bar"
}
有三个实质上不同的地方出现了不重要的筑巢:
这三种方法都需要不同的方法#1在这个问题上被观察到.
要解决#2或#3，请参见Is there a way to omit wrapper/root objects when deserializing objects with Serde?