Rails hash null检查- 'if-condition‘和'ternary operator’哪个更好？

Question

我有下面的散列

instance_options[:param_page]
instance_options[:param_page][:comment_page]
instance_options[:param_page][:comment_per]

我需要comment_page，comment_per值我做了重构代码，但我不确定什么是最好的哈希空检查让我知道最好的空检查，如果你知道其他人

在此之前

comment_page = 1
comment_per = 0
param_page = instance_options[:param_page]
if param_page
  comment_page = param_page[:comment_page].presence || comment_page
  comment_per = param_page[:comment_per].presence || comment_per
end

之后

comment_page = instance_options[:param_page][:comment_page].present? ? instance_options[:param_page][:comment_page] : 1
comment_per = instance_options[:param_page][:comment_per].present? ? instance_options[:param_page][:comment_per] : 10

Answer 1

您可以更进一步：

comment_page = instance_options.dig(:param_page, :comment_page).presence || 1
comment_per  = instance_options.dig(:param_page, :comment_per).presence || 0

这将对instance_options进行两次挖掘，但成本将微不足道。如果您需要考虑类型(假设:comment_page值可以是一个字符串)，那么您可以添加#to_i调用：

comment_page = (instance_options.dig(:param_page, :comment_page).presence || 1).to_i
comment_per  = (instance_options.dig(:param_page, :comment_per).presence || 0).to_i

或者使用reverse_merge (因为这是用于设置默认值)：

param_page = instance_options[:param_page] || {}
param_page.reverse_merge(comment_page: 1, comment_per: 0)
# Then use param_page[:comment_page] and param_page[:comment_per] instead of two locals