# HW4

## 文件議題

* 取得一個list只有name，且不重複並排序的資料
* [Bill,Brain,KZ]

    @GetMapping()
    public List<String> getAllUserWithoutRepeatedName() {
        List<User> response = this.userService.getAllUser();
        List<String> distinct = response.stream().map(User :: getName).distinct().sorted().collect(Collectors.toList());
        return distinct;
    }

* 取得一個 map，其 key 為 ID；value 為 name
* 1:"Bill"
* 2:"Brian"

    @GetMapping()
    public Map<Integer, String> getAllUserWithMap() {
        List<User> response = this.userService.getAllUser();
        Map<Integer, String> map =
            response.stream().collect(Collectors.toMap(User :: getId, User :: getName));
        return map;
    }

* 取得第一筆 name = KZ 的資料

    @GetMapping("/{name}")
    public User getFirstUserByName(@PathVariable String name) {
        List<User> response = this.userService.getAllUser();
        List<User> result = response.stream().filter(x -> name.equals(x.getName())).collect(Collectors.toList());
        Optional<User> first = result.stream().findFirst();
        if(first.isPresent()){
            User user = first.get();
            return user;
        }
        else{
            return null;
        }
    }

* 將資料先依據 age 排序，再依據 ID 排序

    @GetMapping()
    public List<User> sortedByAgeAndId() {
        List<User> response = this.userService.getAllUser();
        List <User> sorted = response.stream().sorted(Comparator.comparing(User :: getAge).thenComparing(User :: getId))
                                              .collect(Collectors.toList());

        return sorted;
    }

* 取得一個 string 為所有資料的 name, age|name, age
* Bill, 13|KZ, 23

    @GetMapping()
    public String getAllUserInString() {
        List<User> response = this.userService.getAllUser();
        String result = response.stream().map(p -> p.getName()+", "+p.getAge()).collect(Collectors.joining("|"));
        return result;
    }