铁锈书第8章-员工管理的文字界面

Question

使用散列映射和向量，创建文本界面，允许用户向公司的部门添加员工姓名。例如，“添加萨利到工程”或“添加Amir到销售”。然后让用户按部门检索所有人员或公司所有人员的列表，并按字母顺序排序。

最近，我非常喜欢学习一些锈蚀知识和阅读“锈书”，并希望从第8章中得到一些关于我对这个挑战的答案的反馈。

我故意把事情做得很简短和简单--我已经从字面上回答了这个问题，从我的测试中可以看出，所有的标准都满足了。

我将这些部门硬编码到hashmap中，并且我没有非常努力地处理边缘情况或不适当的输入。用户还必须输入非常笨重的命令和准确的措辞。

该界面并不是超级用户友好的，因为在每个命令之后，循环会重新启动，每次都会显示欢迎消息。

请随时提供上述任何一个反馈，但要知道，我至少意识到了这些问题。

我最关心的是惯用的锈蚀和代码结构。

是否有比if / else if / else更好的方法来处理各种情况？

为了可读性和我自己的理智，我不想在主函数中嵌套太多的循环。如果我想让UX变得更加无缝，那么为可能被反复调用的每个命令创建一个函数，比如添加员工、查看部门以及内部循环，会有意义吗？

还有什么能突出我做错了什么吗？

有什么我做得对的吗？

use std::collections::HashMap;
use std::io;

fn main() {
    let mut company: HashMap<String, Vec<String>> = HashMap::new();

    company.insert("engineering".to_string(), Vec::new());
    company.insert("operations".to_string(), Vec::new());
    company.insert("sales".to_string(), Vec::new());

    loop {
        println!("Welcome! What would you like to do?\n 
                  To add employee to a department, type 'Add <employee name> to <department name> (engineering, operations, or sales),\n 
                  To view employees in a department, type 'View department <department name>',\n
                  To view all employees by department, type 'View all',
                  To exit, type 'exit'.
        ");

        let mut entry = String::new();
        io::stdin()
            .read_line(&mut entry)
            .expect("Failed to read line");

        let entry: &str = entry.trim();
        let mut split = entry.split_whitespace();

        if entry.contains("Add") {
            let emp = split.nth(1).unwrap().to_string();
            let dep = split.nth(1).unwrap().to_lowercase();
            println!("{emp} added to {dep}");

            let vec = company.get_mut(&dep).unwrap();
            vec.push(emp);
            vec.sort_by_key(|name| name.to_lowercase());
            println!("{:?}", company)
        } else if entry.contains("View department") {
            let dep = split.nth(2).unwrap().to_lowercase();
            let view = &company[&dep];

            println!("Employees in {dep}:\n");
            for v in view {
                println!("{v}\n");
            }
        } else if entry.contains("View all") {
            for k in company.keys() {
                let view = &company[k];
                println!("Department: {k}");
                for v in view {
                    println!("{v}\n");
                }
            }
        } else if entry == "exit" {
            break;
        } else {
            println!("Please enter a valid option");
        }
    }
}

Answer 1

它已经很好了，但有几件事我会改变。首先，我介绍了使main变得更短的不同方法。

fn view_department(company: &HashMap<String, Vec<String>>, department: &str) {
    println!("Employees in {department}:\n");
    for e in &company[department] {
        println!("{e}\n");
    }
}

fn view_all(company: &HashMap<String, Vec<String>>) {
    for (department, employees) in company {
        println!("Department: {department}");
        for e in employees {
            println!("{e}\n");
        }
    }
}

fn add_employee(company: &mut HashMap<String, Vec<String>>, employee: &str, department: &str) {
    println!("{employee} added to {department}");

    let vec = company.get_mut(department).unwrap();
    vec.push(employee.to_owned());
    vec.sort_by_key(|name| name.to_lowercase());
    println!("{company:?}")
}

注意，我直接迭代hashmap的键值对，这比两次散列略高一些。

您可以简单地使用一个多行字符串来消除换行符：

        println!(
"Welcome! What would you like to do?
To add employee to a department, type 'Add <employee name> to <department name> (engineering, operations, or sales),
To view employees in a department, type 'View department <department name>',
To view all employees by department, type 'View all',
To exit, type 'exit'."
        );

最重要的是，由于问题在于不同字符串模式的匹配，因此我使用模式匹配来实现：

match entry.split_whitespace().collect::<Vec<&str>>().as_slice() {
    ["Add", emp, "to", dep] => add_employee(&mut company, emp, dep),
    ["View", "department", dep] => view_department(&company, dep),
    ["View", "all"] => view_all(&company),
    ["exit"] => break,
    _ => println!("Please enter a valid option")
}

我认为这比包含大量nth和神奇索引的版本更易读。

附带注意:我最初写的是entry.trim().split_whitespace()，但是告诉我trim是不必要的，所以经常使用clippy，并听取它的建议。

以下是供参考的全部主要内容：

fn main() {
    let mut company: HashMap<String, Vec<String>> = HashMap::new();

    company.insert("engineering".to_string(), Vec::new());
    company.insert("operations".to_string(), Vec::new());
    company.insert("sales".to_string(), Vec::new());

    loop {
        println!(
"Welcome! What would you like to do?
To add employee to a department, type 'Add <employee name> to <department name> (engineering, operations, or sales),
To view employees in a department, type 'View department <department name>',
To view all employees by department, type 'View all',
To exit, type 'exit'."
        );

        let mut entry = String::new();
        io::stdin()
            .read_line(&mut entry)
            .expect("Failed to read line");

        match entry.split_whitespace().collect::<Vec<&str>>().as_slice() {
            ["Add", emp, "to", dep] => add_employee(&mut company, emp, dep),
            ["View", "department", dep] => view_department(&company, dep),
            ["View", "all"] => view_all(&company),
            ["exit"] => break,
            _ => println!("Please enter a valid option")
        }
    }
}
```