尽可能地抽象，我想执行：

select id,complicatedCalculation as geodistance from User join others having geodistance < 100;

在CakePHP 2.x中。实际的查询使用聚合，所以我们确实需要 having。当数据库是MySQL时，这一切都可以工作，因为该数据库在 having 子句中接受列引用。我们现在转向Postgres，它更严格地实现了SQL标准，并报告geodistance不存在。
SQL中的解决方案是将指令编写为

select * from (select id, complicatedCalculation as geodistance from User join others) as sub where geodistance < 100;

但是我不知道如何在Cake中表达这个子查询。目前为止我最好的尝试是

$db = $this->User->getDataSource();
$core = [ 'fields' => [ 'id', 'complicatedCalculation as geodistance' ],
        'table' => $db->fullTableName($this->User),
        'join' => (others) ]
$sub_query = $db->buildStatement($core, $this->User);
$result = $this->User->find('all', [
        'fields' => [ 'id', 'geodistance' ],
        'table' => $db->expression($sub_query),
        'alias' => 'zzz',  // not used but PGSQL requires it.
        'conditions' => [ 'geodistance <=' => 100 ]
    ]);

但这会在日志中生成以下SQL（抽象）报告为错误：

SELECT "User"."id" AS "User__id", complicatedCalculation as geodistance FROM "public"."users" AS "User" LEFT JOIN (others) WHERE "geodistance" <= 100

所以我尝试让CakePHP使用表达式作为表是不起作用的。我能让它这么做吗？
建议升级到当前版本的Cake的答案是不可接受的。如果没有其他方法，我们就使用query()。