如何解析 JSON 文件?

How do I parse a JSON File?

到目前为止,我的目标是在 Rust 中解析这个 JSON 数据:

extern crate rustc_serialize;
use rustc_serialize::json::Json;
use std::fs::File;
use std::io::copy;
use std::io::stdout;

fn main() {
    let mut file = File::open("text.json").unwrap();
    let mut stdout = stdout();
    let mut str = &copy(&mut file, &mut stdout).unwrap().to_string();
    let data = Json::from_str(str).unwrap();
}

并且text.json

{
    "FirstName": "John",
    "LastName": "Doe",
    "Age": 43,
    "Address": {
        "Street": "Downing Street 10",
        "City": "London",
        "Country": "Great Britain"
    },
    "PhoneNumbers": [
        "+44 1234567",
        "+44 2345678"
    ]
}

解析它的下一步应该是什么?我的主要目标是获取这样的 JSON 数据,并从中解析一个键,例如 Age.

由 Rust 社区的许多有用成员解决:

extern crate rustc_serialize;
use rustc_serialize::json::Json;
use std::fs::File;
use std::io::Read;

fn main() {
    let mut file = File::open("text.json").unwrap();
    let mut data = String::new();
    file.read_to_string(&mut data).unwrap();

    let json = Json::from_str(&data).unwrap();
    println!("{}", json.find_path(&["Address", "Street"]).unwrap());
}

Serde 是首选 JSON 序列化提供程序。你可以 . Once you have it as a string, use serde_json::from_str:

fn main() {
    let the_file = r#"{
        "FirstName": "John",
        "LastName": "Doe",
        "Age": 43,
        "Address": {
            "Street": "Downing Street 10",
            "City": "London",
            "Country": "Great Britain"
        },
        "PhoneNumbers": [
            "+44 1234567",
            "+44 2345678"
        ]
    }"#;

    let json: serde_json::Value =
        serde_json::from_str(the_file).expect("JSON was not well-formatted");
}

Cargo.toml:

[dependencies]
serde = { version = "1.0.104", features = ["derive"] }
serde_json = "1.0.48"

您甚至可以使用 serde_json::from_reader 之类的东西直接从打开的 File 中读取。

Serde 可用于 JSON 以外的格式,它可以序列化和反序列化为自定义结构而不是任意集合:

use serde::Deserialize;

#[derive(Debug, Deserialize)]
#[serde(rename_all = "PascalCase")]
struct Person {
    first_name: String,
    last_name: String,
    age: u8,
    address: Address,
    phone_numbers: Vec<String>,
}

#[derive(Debug, Deserialize)]
#[serde(rename_all = "PascalCase")]
struct Address {
    street: String,
    city: String,
    country: String,
}

fn main() {
    let the_file = /* ... */;

    let person: Person = serde_json::from_str(the_file).expect("JSON was not well-formatted");
    println!("{:?}", person)
}

查看 Serde website 了解更多详情。

赞成已接受的答案(因为它有帮助),但只是添加我的答案,使用@FrickeFresh

引用的广泛使用的serde_json crate

假设您的foo.json

{
    "name": "Jane",
    "age": 11
}

实现看起来像

extern crate serde;
extern crate json_serde;
#[macro_use] extern crate json_derive;
use std::fs::File;
use std::io::Read;

#[derive(Serialize, Deserialize)]
struct Foo {
    name: String,
    age: u32,
}

fn main() {
   let mut file = File::open("foo.json").unwrap();
   let mut buff = String::new();
   file.read_to_string(&mut buff).unwrap();

   let foo: Foo = serde_json::from_str(&buff).unwrap();
   println!("Name: {}", foo.name);
}

serde_json::de::from_reader 文档中有一个简短而完整的示例,说明如何从文件中读取 JSON。

这是一个简短的片段:

  • 正在读取文件
  • 将其内容解析为 JSON
  • 并使用所需的键提取字段

享受:

let file = fs::File::open("text.json")
    .expect("file should open read only");
let json: serde_json::Value = serde_json::from_reader(file)
    .expect("file should be proper JSON");
let first_name = json.get("FirstName")
    .expect("file should have FirstName key");