简体   繁体   English

Ruby on Rails-访问当前用户的嵌套属性

[英]Ruby on Rails - accessing nested attributes of current user

i am trying to access some nested attributes in a rails partial; 我正在尝试访问部分Rails中的嵌套属性; let me explain my issue: 让我解释一下我的问题:

i wrote a simple messaging service according to the tutorial of dana mulder i mentioned in another question of mine. 我根据我在另一个问题中提到的dana mulder的教程编写了一个简单的消息服务。 a user has_many conversations , and a conversation has_many messages . user has_many conversations ,而conversation has_many messages every conversation has a recipient and a sender and every message has a boolean read , which is set to false by default. 每个对话都有一个收件人和一个发件人,每个消息都有一个布尔型read ,默认情况下将其设置为false

what i am trying to do is to write a simple function that iterates through every message a user got in a conversation he is involved in, checks if the last message a user recieved in a conversation is read == true and not send by himself to display a little knob beside the "messages" link in the navigation of my app. 我要做的是编写一个简单的函数,该函数遍历用户所参与的对话中的每条消息,检查用户在对话中收到的最后一条消息是否读为== true,而不是自己发送给在我的应用程序导航中的“消息”链接旁边显示一个小旋钮。

What i basically want to check is (multiple lines for better readability): 我基本上要检查的是(多行以提高可读性):

<%= if current_user.conversations.all.messages.last.user_id != current_user.id && current_user.conversations.all.messages.last.read == false %>

    Message link with bubble beside it

<% else %>

    Message link

<% end %>

This syntax does not work .. How is it possible to iterate through every conversation a user is involved, check the last message of every conversation if its written by this user and if not if its already read? 此语法不起作用..如何遍历用户所涉及的每个会话,检查每个会话的最后一条消息(如果此用户写的消息,如果不是已读的消息)?

while something like Conversation.last.messages.last.content is working, current_user.conversations.all.messages.last.content is not.. i find it a bit confusing how the accessibility of rails models are working. 虽然像Conversation.last.messages.last.content这样的东西正在工作,但current_user.conversations.all.messages.last.content却没有..我发现它有点困扰Rails模型的可访问性。

thanks in advance! 提前致谢! i hope i was explaining me well enough. 我希望我对我的解释足够好。

Best Regards 最好的祝福

What are you trying to do with current_user.conversations.all.messages.last.content ? 您要如何处理current_user.conversations.all.messages.last.content because conversations.all is an association, it is NOT a Conversation, so messages is not an available method. 因为conversations.all是关联,所以它不是对话,因此messages不是可用的方法。 Conversation.last IS a Conversation, that's why you can call messages on it. Conversation.last是一种对话,因此您可以在其上调用messages

You can try current_user.conversations.last.messages.last.content to get the last message of the last conversation, or, you can use a "has_many :through" relationship on User 您可以尝试使用current_user.conversations.last.messages.last.content来获取上次对话的最后一条消息,也可以在User上使用“ has_many:through”关系

class User ...
  has_many :conversations
  has_many :messages, through: :conversations
end

That way you can do current_user.messages if you want ALL user's messages and current_user.messages.last to get the last Message (even if it's not from the last conversation). 这样,如果您希望所有用户的消息和current_user.messages.last都获得最后一条消息(即使它不是来自上一次对话),则可以执行current_user.messages

I'd recommend you to read the Rails Guide about Associations https://guides.rubyonrails.org/association_basics.html#the-has-many-through-association 我建议您阅读有关关联的Rails指南https://guides.rubyonrails.org/association_basics.html#the-has-many-through-association

EDIT: To know if the user has some unread message, you can use the has_many :through association and do something like 编辑:要知道用户是否有未读消息,可以使用has_many :through关联并执行类似的操作

current_user.messages.where(read: false).where.not(user_id: current_user.id).any?

That will return true any message from the user's conversations with id != current_user.id is not read. 这将返回true则不会读取用户对话中ID为!= current_user.id的任何消息。 And false otherwise. 否则为false

You can use count instead of any? 您可以使用count代替any? if you wan't to know the actual number of unread messages. 如果您不想知道未读邮件的实际数量。

声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM