如果您设置的关联正确实在是小巫见大巫:
class User < ApplicationRecord
has_many :recieved_messages, class_name: 'Message', foreign_key: 'recipent_id'
has_many :sent_messages, class_name: 'Message', foreign_key: 'sender_id'
def all_messages
Message.where('recipient_id = :id OR sender_id = :id', id: id)
end
end
class Message < ApplicationRecord
belongs_to :recipient, class_name: 'User', inverse_of: :recieved_messages
belongs_to :sender, class_name: 'User', inverse_of: :sent_messages
def self.between(a,b)
.where('(recipient_id = :id OR sender_id = :id)', id: a)
.where('(recipient_id = :id OR sender_id = :id)', id: b)
end
end
# all messages recieved by the current user:
current_user.recieved_messages.order(created_at: :desc)
# all messages recieved by the current user:
current_user.sent_messages.order(created_at: :desc)
# all messages sent or recieved by the current user:
current_user.all_messages.order(created_at: :desc)
# all messages between two users
Message.between(current_user, some_other_user)
,这是让用户之间的任何消息的关键是:
WHERE (recipient_id = a OR sender_id = a) AND (recipient_id = b OR sender_id = b)
但你应该考虑创建一个会话模型来加入它们。
调用.last
将按降序排列后得到最早的记录。如果你想要最新的消息使用.first
为了避免N + 1查询问题,请确保您加载相关的记录:
@messages.eager_load(:recipient, :sender)
当迭代(一个一般)使用的关联,而不是IDS。千万别像User.find(message.to_id)
这样会造成N + 1查询问题。
<% @messages.each do |message| %>
<p><b><%= message.recipent.name %></b></p>
<p>
<% if message.read? %>
<%= link_to message.content, pm_path(message.recipient) %>
<% else %>
<b><%= link_to message.content, pm_path(message.recipient) %></b>
<% end %>
</p>
<% end %>
来源
2017-10-14 10:59:11
max
Dosent work .... –