如何在 Sequel Ruby 模型中获取一个表中而不是另一个表中的所有记录

发布于 2025-01-09 22:26:42 字数 592 浏览 0 评论 0原文

这个想法是返回 table_1 模型的模型实例，其中 table_1 的 id 不存在于 table_2 中，而 another_id 具有值。这样我就可以使用这些实例和其他地方包含的数据

def self.my_method(another_id:)
  result = DB["SELECT t1.* FROM table_1 t1 WHERE t1.id NOT IN (SELECT t2.id 
  FROM table_2 t2 WHERE t2.another_id = #{another_id})"]

  result_array = []
  result.each do |row| 
   result_array << self.find(id: row[:id])
  end

 result_array
end

我有 table_1 和 table_2 的模型

这可以工作，但感觉非常笨重并且可能非常低效，因为对于返回的每个 id 我正在重建每个 table_1 模型的实例再次查看数据库。

如何使用模型/数据集上可用的方法以典型的 Sequel 方式完成此操作。从我的角度来看，我正在努力解决如何将各种语法组合在一起来做到这一点

原文

The idea is to return model instances of the table_1 model where the id's of table_1 do not exist in table_2 where another_id has a value. This is so I can use those instances and the data contained within elsewhere

def self.my_method(another_id:)
  result = DB["SELECT t1.* FROM table_1 t1 WHERE t1.id NOT IN (SELECT t2.id 
  FROM table_2 t2 WHERE t2.another_id = #{another_id})"]

  result_array = []
  result.each do |row| 
   result_array << self.find(id: row[:id])
  end

 result_array
end

I have models for both table_1 and table_2

This works but it feels very clunky and probably very inefficient as for every id that is returned I am rebuilding an instance of each table_1 model by looking at the database again.

How would this be done in a typically Sequel way using the methods available on the Model/Dataset. From my viewpoint I am struggling with how I pull together the various pieces of syntax to do this

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