將複雜查詢轉換爲cakephp分頁條件

Question

我允許我的用戶選擇英里半徑數字和位置，然後我試圖獲取位於該半徑內的結果。有一羣在線教程，並有一個這似乎有完善的查詢

'SELECT id, (3959 * acos(cos(radians(37)) * cos(radians(lat)) * cos(radians(lng) - radians(-122)) + sin(radians(37)) * sin(radians(lat)))) AS distance FROM markers HAVING distance < 25 ORDER BY distance LIMIT 0 , 20;' 

但我使用CakePHP，我想使用分頁組件，所以我需要把這種查詢到條件陣列選擇所有標記和相關數據。誰能幫我嗎？我是cakephp的新手，在最好的時候絕對沒有用sql。

特別的HAVING位彆扭......

編輯

我現在使用此代碼

$this->Paginator->settings = array(
      'fields'=>array('id','(3959 * acos(cos(radians(37)) * cos(radians('.$lat.')) * cos(radians('.$lng.') - radians(-122)) + sin(radians(37)) * sin(radians('.$lat.')))) AS distance '), 
           'limit' => 20, 
           'conditions'=>array('distance < '.$miles) 
     ); 
$data = $this->Paginator->paginate('Job'); 
$this->set(compact('data')); 

並且至少正在運行的查詢，但我得到錯誤說

Unknown column 'distance' in 'where clause' 

任何想法？

更多修改

的日誌中顯示

SELECT `Job`.`id`, (3959 * acos(cos(radians(37)) * cos(radians(54.59728500000001)) * cos(radians(-5.930119999999988) - radians(-122)) + sin(radians(37)) * sin(radians(54.59728500000001)))) AS distance FROM `trainnation`.`jobs` AS `Job` LEFT JOIN `trainnation`.`users` AS `User` ON (`Job`.`user_id` = `User`.`id`) LEFT JOIN `trainnation`.`courses` AS `Course` ON (`Job`.`course_id` = `Course`.`id`) WHERE 1 = 1 GROUP BY distance having distance < 49 LIMIT 20 

看起來正確的，但它不是從DB現在拉任何東西。而且應該從這些選項中的至少9

Answer 1

試試這個

$this->paginate = array('fields'=>array('id','(3959 * acos(cos(radians(37)) * cos(radians(lat)) * cos(radians(lng) - radians(-122)) + sin(radians(37)) * sin(radians(lat)))) AS distance '), 
'limit' => 20, 
'group'=>'distance having distance < 25 ') 
); 
$marker_data = $this->paginate('Marker');