用服装

t = Post.arel_table

results = Post.where(
  t[:author].eq("Someone").
  or(t[:title].matches("%something%"))
)

结果SQL:

ree-1.8.7-2010.02 > puts Post.where(t[:author].eq("Someone").or(t[:title].matches("%something%"))).to_sql
SELECT     "posts".* FROM       "posts"  WHERE     (("posts"."author" = 'Someone' OR "posts"."title" LIKE '%something%'))

rails + arel，一个更清晰的方式:

# Table name: messages
#
# sender_id:    integer
# recipient_id: integer
# content:      text

class Message < ActiveRecord::Base
  scope :by_participant, ->(user_id) do
    left  = arel_table[:sender_id].eq(user_id)
    right = arel_table[:recipient_id].eq(user_id)

    where(Arel::Nodes::Or.new(left, right))
  end
end

生产:

$ Message.by_participant(User.first.id).to_sql 
=> SELECT `messages`.* 
     FROM `messages` 
    WHERE `messages`.`sender_id` = 1 
       OR `messages`.`recipient_id` = 1

只要在条件中添加一个OR

Model.find(:all, :conditions => ["column = ? OR other_column = ?",value, other_value])

Rails 5附带了一个or方法。(链接到文档)

该方法接受ActiveRecord::Relation对象。例如:

User.where(first_name: 'James').or(User.where(last_name: 'Scott'))

用服装

t = Post.arel_table

results = Post.where(
  t[:author].eq("Someone").
  or(t[:title].matches("%something%"))
)

结果SQL:

ree-1.8.7-2010.02 > puts Post.where(t[:author].eq("Someone").or(t[:title].matches("%something%"))).to_sql
SELECT     "posts".* FROM       "posts"  WHERE     (("posts"."author" = 'Someone' OR "posts"."title" LIKE '%something%'))

你可以这样做:

Person.where("name = ? OR age = ?", 'Pearl', 24)

或者更优雅一点，安装rails_or gem并像这样做:

Person.where(:name => 'Pearl').or(:age => 24)