我想在层次结构深处的对象中打印每个联系人的名称.每次制作合适的结构时,接触对象可能不具有完全相同数量的字段.我怎样才能做到这一点?
extern crate serde_json;
use serde_json::{Error, Value};
use std::collections::HashMap;
fn untyped_example() -> Result<(), Error> {
// Some JSON input data as a &str. Maybe this comes from the user.
let data = r#"{
"name":"John Doe",
"age":43,
"phones":[
"+44 1234567",
"+44 2345678"
],
"contact":{
"name":"Stefan",
"age":23,
"optionalfield":"dummy field",
"phones":[
"12123",
"345346"
],
"contact":{
"name":"James",
"age":34,
"phones":[
"23425",
"98734"
]
}
}
}"#;
let mut d: HashMap<String, Value> = serde_json::from_str(&data)?;
for (str, val) in d {
println!("{}", str);
if str == "contact" {
d = serde_json::from_value(val)?;
}
}
Ok(())
}
fn main() {
untyped_example().unwrap();
}
我是Rust的新手,基本上来自JavaScript背景.
最佳答案
may not have the exact same number of fields every time to make a
suitable structure
目前还不清楚你为什么这么想:
extern crate serde_json; // 1.0.24
#[macro_use]
extern crate serde_derive; // 1.0.70;
use serde_json::Error;
#[derive(Debug, Deserialize)]
struct Contact {
name: String,
contact: Option<Box<Contact>>,
}
impl Contact {
fn print_names(&self) {
println!("{}", self.name);
let mut current = self;
while let Some(ref c) = current.contact {
println!("{}", c.name);
current = &c;
}
}
}
fn main() -> Result<(), Error> {
let data = r#"{
"name":"John Doe",
"contact":{
"name":"Stefan",
"contact":{
"name":"James"
}
}
}"#;
let person: Contact = serde_json::from_str(data)?;
person.print_names();
Ok(())
}
我已经从JSON中删除了额外的字段以获得一个较小的示例,但如果它们存在则没有任何变化.
也可以看看:
> Learning Rust With Entirely Too Many Linked Lists
> Why are recursive struct types illegal in Rust?
> Rust & Serde JSON deserialization examples?