Question

升级Rails 3.2。到Rails 4.我有以下范围：

# Rails 3.2
scope :by_post_status, lambda { |post_status| where("post_status = ?", post_status) }
scope :published, by_post_status("public")
scope :draft, by_post_status("draft")

# Rails 4.1.0
scope :by_post_status, -> (post_status) { where('post_status = ?', post_status) }

但我无法找到如何做第2和第3行。如何从第一个范围创建另一个范围？

Answer 1

非常简单，只是没有参数的lambda：

scope :by_post_status, -> (post_status) { where('post_status = ?', post_status) }
scope :published, -> { by_post_status("public") }
scope :draft, -> { by_post_status("draft") }

或更短的：

%i[published draft].each do |type|
  scope type, -> { by_post_status(type.to_s) }
end

Answer 2

来自Rails edge docs

＆＃34; Rails 4.0要求范围使用可调用对象，例如Proc或lambda：＆＃34;

scope :active, where(active: true)

# becomes 
scope :active, -> { where active: true }

考虑到这一点，您可以轻松地重写代码：

scope :by_post_status, lambda { |post_status| where('post_status = ?', post_status) }
scope :published, lambda { by_post_status("public") }
scope :draft, lambda { by_post_status("draft") }

如果您希望支持许多不同的状态，并认为这很麻烦，以下内容可能适合您：

post_statuses = %I[public draft private published ...]
scope :by_post_status, -> (post_status) { where('post_status = ?', post_status) }

post_statuses.each {|s| scope s, -> {by_post_status(s.to_s)} }

Rails 4范围与参数

2 个答案: