如何在两个或多个实例之间共享JOIN的结果

为了减少查询的数量，当我从数据库中选择一些东西时，我想要启动最好的查询，它可以是以下之一:

1- SELECT ... FROM cities JOIN states ON city.stateID = state.stateID WHERE city.cityID = ...
2- SELECT ... FROM cities WHERE city.cityID = ...
3- SELECT ... FROM states WHERE state.stateID = ...

如果我执行第一个查询，我将不需要(可能)执行第二个或第三个查询，因为我已经有数据。

但是如何在两个或多个实例类之间共享单个连接查询的结果?假设我必须首先使用第3个查询，因为我已经有了来自3的数据，您如何从代码中控制City实例的创建?

<?php
 class City
 {
      protected $state;
      function getState()
      {
           if(!$this->state)
                 $this->state = State::getByID($this->stateID);
           return $this->state;
      }
      // since i may already have (in the outer scope) the $state instance...
      // to avoid an unnecessary query (State::getByID()) i allow to set the state:
      function setState(State $state)
      {
           if($state->stateID == $this->stateID)
                $this->state = $state;
           else
                throw new Exception("This City doesn't belong to the given State");
      }
 }

是正确的吗?我做得对吗?

<?php
class Address
{
    const PREFETCH_CITY = 1;
    const PREFETCH_STATE = 2;
    const PREFETCH_ALL = 3;
    construct($cityID, $prefetch = 0)
    {
         if($prefetch & static::PREFETCH_CITY)
              // i use the "JOIN cities" query (i also set $this->city)
         elseif($prefetch & static::PREFETCH_STATE)
              // i use the "JOIN states" query (i also set $this->state)
         elseif($prefetch & static::PREFETCH_ALL)
              // i use the "JOIN states JOIN cities" query
              // (i preset also both $this->city and $this->state)
         else
              // i don't use any JOIN, i just get the city
    }
}

实际上现在我认为构造函数应该在一个单独的类，但无论如何…

一般说话……关于这个论点我能读到什么?欢迎阅读书籍、教程

希望我已经说清楚了，我的英语很糟糕…非常感谢你提前:)