在使用serde_json解析时,是否可以展平子对象字段?
在使用serde_json解析时,是否可以展平子对象字段?,json,rust,serde,Json,Rust,Serde,#[serde(rename)]似乎是正确的选择,但文档没有说明是否可行或如何实现 此JSON对象: { “名称”:“myobject” “信息”: { “计数器”:“3” “foo”:“bar” } } 相应的平锈结构应为: #[derive(Deserialize)] struct Object { name: String, #[serde(rename="info.counter")] // wrong syntax here !! count: i32,
#[serde(rename)]
似乎是正确的选择,但文档没有说明是否可行或如何实现
此JSON对象:
{
“名称”:“myobject”
“信息”:
{
“计数器”:“3”
“foo”:“bar”
}
}
相应的平锈结构应为:
#[derive(Deserialize)]
struct Object {
name: String,
#[serde(rename="info.counter")] // wrong syntax here !!
count: i32,
#[serde(rename="info::foo")] // neither this works
foo: String,
}
没有内置的方法来处理属性,但是您可以为您的
对象
类型编写自己的反序列化
impl,该类型首先反序列化到某个中间帮助器表示,然后将数据重新排列到所需的结构中
#[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());
}
有三个实质上不同的地方出现了不重要的嵌套:
Object {
name: "myobject",
count: 3,
foo: "bar"
}