Question

我将我的应用与Code Climate同步进行代码审核。

LIB /模型/用户/同步/ basecamp_inc.rb

def todo_list(todos,basecamp_account,proj_id)
  todos.map{|todo|
    task = basecamp_tasklist(proj_id,todo['id'],basecamp_account)              
    task_list(task['todos']['remaining'],task['name'],'remain') if task['todos']['remaining'].present?
    task_list(task['todos']['completed'],task['name'],'complete') if task['todos']['completed'].present?
  }
end

def task_list(tasks,todo_list,status) 
  tasks.map{|task|
    task_created(task,todo_list)
    task['assignee'].present? && user_owner?(task['assignee']['name']) ? save_task_activity('was assigned to a',task,todo_list,task['updated_at'],'assign') : [tasks,todo_list,status]
    status == 'complete' && user_owner?(task['completer']['name']) && current_date?(task['completed_at']) ? save_task_activity('finished',task,todo_list,task['completed_at'],'finished') : [tasks,todo_list,status]          
  }
end

上述代码被Code Climate检测为具有复杂方法。我不明白为什么Code Climate认为它很复杂。那么，该代码的等级仍然是C，我想重构它。我怎样才能改善这个？

Answer 1

您应该考虑将以下条件转换为私有方法：

task['assignee'].present? && user_owner?(task['assignee']['name']) ? save_task_activity('was assigned to a',task,todo_list,task['updated_at'],'assign') : [tasks,todo_list,status]
status == 'complete' && user_owner?(task['completer']['name']) && current_date?(task['completed_at']) ? save_task_activity('finished',task,todo_list,task['completed_at'],'finished') : [tasks,todo_list,status]

您应该考虑将以下调用转移到采用status参数的单个方法，例如：

task_list(task['todos']['remaining'], task['name'], 'remain') if task['todos']['remaining'].present?

将被重构为：

def build_task_list(tasks, name, status)
  task_list(tasks, name, status) if task.present?
end

build_task_list(task['todos']['remaining'], task['name'], 'remain')
build_task_list(task['todos']['completed'], task['name'], 'complete')

RAILS中的重构方法复杂性

1 个答案: