code samples from https://news.ycombinator.com/item?id=22712441 with updates

problem.md

Ok, let's say you have this json:

[{
    "name": "Kévin",
    "age": 23,
    "hired": "2005-06-03 02:12:33",
    "emails": ["kevin@foo.com", "kevin@bar.com"]
}, {
}, {
    "name": "Joël",
    "age": 32,
    "hired": "2003-01-02 12:32:11",
    "emails": ["joel@foo.com", "joel@bar.com"]
},
... other entries
]

It's very simple. Very basic. There is no trick in there: it's standard utf8, well formed, no missing value. You want to print people details in alphabetical order this way:

Joël (32) - 02/01/03:
- joel@foo.com
- joel@bar.com
Kévin (23) - 03/06/05:
- kevin@foo.com
- kevin@bar.com
... other entries

solution.py

    import json
    import datetime as dt

    with open("agenda.json") as fd:

        agenda = sorted(json.load(fd), key=lambda people: people["name"])

        for people in agenda:

            hired = dt.datetime.fromisoformat(people["hired"])
            print(f'{people["name"]} ({people["age"]}) - {hired:%d/%m/%y}:')

            for email in people["emails"]:
                print(f" - {email}")

solution.rs

use std::fs;

use anyhow::Result;
use chrono::NaiveDateTime;
use serde::*;
use serde_json;


#[derive(Deserialize)]
struct Person {
    name: String,
    age: u32,
    hired: String,
    emails: Vec<String>,
}

fn main() -> Result<()> {
    let data = fs::read_to_string("agenda.json")?;
    let mut people: Vec<Person> = serde_json::from_str(&data)?;

    people.sort_by(|a, b| b.name.cmp(&a.name));

    for person in &people {
        let datetime = NaiveDateTime::parse_from_str(&person.hired, "%Y-%m-%d %H:%M:%S")?;
        println!(
            "{} ({}) - {}",
            person.name,
            person.age,
            datetime.format("%d/%m/%y")
        );

        for email in &person.emails {
            println!(" - {}", email);
        }
    }
    
    Ok(())
}