CakePHP和子查詢

Question

如何使用Cake語法編寫SQL子查詢？ 我知道如何編寫簡單的查詢，但我無法處理子查詢。

這是原始查詢：

SELECT Assumption.id, Referee.id, Referee.first_name, Referee.second_name 
FROM referees AS Referee 
INNER JOIN (

    SELECT a.id, a.referee_id 
    FROM assumptions a 
    WHERE a.season_id =7 
) AS Assumption ON Referee.id = Assumption.referee_id 

Answer 1

既然你聽不懂的語法，這是實際的查詢：

$records = $this->Referee->find('all', array(
       'fields' => array(
        'Assumption.id', 'Referee.id', 'Referee.first_name', 'Referee.second_name' 
        ), 
       'joins' => array(
        array(
         'table' => 'assumptions', 
         'alias' => 'Assumption', 
         'type' => 'INNER', 
         'foreignKey' => false, 
         'conditions' => array('Referee.id = Assumption.referee_id', 'Assumption.season_id = 7'), 
         ), 
       ), 
      ) 
     ); 

將會產生這個查詢：

SELECT 
    `Assumption`.`id`, 
    `Referee`.`id`, 
    `Referee`.`first_name`, 
    `Referee`.`second_name` 
FROM `referees` AS `Referee` 
INNER JOIN assumptions AS `Assumption` 
    ON (`Referee`.`id` = `Assumption`.`referee_id` 
     AND `Assumption`.`season_id` = 7) 

其中提供您正在尋找的結果。

輸出示例：

Array 
(
    [0] => Array 
     (
      [Assumption] => Array 
       (
        [id] => 1 
        [0] => Array 
         (
          [id] => 1 
          [season_id] => 7 
          [referee_id] => 1 
          [name] => SomeAssumpton 
         ) 

       ) 

      [Referee] => Array 
       (
        [id] => 1 
        [first_name] => Ref 
        [second_name] => one 
       ) 

     ) 

) 

Answer 2

打完似乎像年（幾個小時），閱讀蛋糕源...如果你想簡單地鍵入您的子查詢到你的蛋糕條件......

它使用PHP stdClass已經不是一個數組項...它只會轉儲你的「價值」進入查詢......

$subquery = new stdClass(); 
    $subquery->type = "expression"; 
    $subquery->value = "Product.id IN (select product_id from product_categories where category_id='$category_id' or belongs_to='$category_id')"; 

    $options['conditions'][] = $subquery; <- dump the class into your conditions array! 

做普通查詢 $這個 - >表 - >找到（「所有」， $選項）

實例：（正常蛋糕，子查詢的quickfix）

//only from my vendor 
    $options['conditions']['collection_id'] = $vendor_id; 

    //inner join to CollectionProduct 
    $options['joins'][0] = array(
     "table" => "collection_products", 
     "alias" => "CollectionProduct", 
     "type" => "INNER", 
     "conditions" => array(
      "Product.id = CollectionProduct.product_id", 
     ), 
    ); 

    //show only from current category 
    if ($category_id) { 
     $subquery = new stdClass(); 
     $subquery->type = "expression"; 
     $subquery->value = "Product.id IN (select product_id from product_categories where category_id='$category_id' or belongs_to='$category_id')"; 
     $options['conditions'][] = $subquery; 
    } else { 
     //get 18 random items... no category selected? 
     $options['limit'] = 18; 
    } 

    return $this->find('all', $options); 

Answer 3

cdburgess的答案是這種情況的正確的;它以最簡單，最直接的方式解決問題。

這就是說，當解決執行子查詢的一般問題時，食譜記錄了preferred solution to the subquery problem。這不是很優雅，但是這是指南所說的做法。

或者你可以非常小心，只要做一個$ this-> Model-> query（）。