CakePHP加入問題

Question

這讓我瘋狂。這不會引發任何錯誤，但它也不會執行連接。我希望這是一個，我已經花了太長時間看着它，答案是顯而易見的給別人...

 $lines = $this->RevenueLine->find('all', array(
       'conditions' => array(
        'RevenueLine.is_triggered' => 1, 
        'RevenueLine.date_triggered >=' => $sqldate1, 
        'RevenueLine.date_triggered <=' => $sqldate2, 
       ), 
       'joins' => array(
        array(
         'table' => 'projects', 
         'alias' => 'Project', 
         'type' => 'INNER', 
         'conditions' => array(
          'RevenueLine.project_id = Project.id' 
         ) 
        ), 
        array(
         'table' => 'clients', 
         'alias' => 'Client', 
         'type' => 'INNER', 
         'conditions' => array(
          'Project.client_id = Client.id' 
         ) 
        ), 
        array(
         'table' => 'classifications', 
         'alias' => 'Classification', 
         'type' => 'INNER', 
         'conditions' => array(
          'Project.classification_id = Classification.id' 
         ) 
        )      
       ), 
       'order' => array(
        'Client.client_number ASC', 
        'Project.pn_counter ASC' 
       ) 
      ) 
     ); 

Answer 1

您需要選擇從連接表中的字段：

'fields' => array(
    'JoinTable1.*', 
    'JoinTable2.*', 
    'JoinTable3.*', 
    'JoinTable4.*' 
) 

作爲您查找的參數。