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 = ©(&mut file, &mut stdout).unwrap().to_string();
let data = Json::from_str(str).unwrap();
}
并且text.json
是
{
"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
是
{
"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");
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 = ©(&mut file, &mut stdout).unwrap().to_string();
let data = Json::from_str(str).unwrap();
}
并且text.json
是
{
"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 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
是
{
"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");