遞歸函數邏輯問題

Question

我有一個遞歸函數，帶有一個我遇到問題的錯誤。邏輯如下： 表單具有各種類型的字段。一種類型的字段是引用字段，它將表單鏈接到另一個表單。而且，參考文獻可以鏈接。例如：我可以有一個處方表格，裏面有一個參考文獻，對醫生有一個提及人的參考。同樣的相遇表格有一個對病人的引用，也指人。

因此，醫生姓氏的參考字段鏈看起來像處方上的字段49指的是遇到。 Encounter上的Field 7是指Doctor。 Doctor上的第6場是指Person。個人領域1指的是姓氏。 [49-7-6-1]

患者姓氏的引用鏈將遵循相同的邏輯[49-5-4-1]。

我的問題是這樣的。使用當前函數，如果我在較低級別分支引用，則引用鏈字符串會有問題。 [49-5-4-1]代替患者姓氏[49-5-4-1]，我得到[49-7-5-4-1]。

功能如下：

function get_reference_fields($form_id, $base_fields = array(), $ref_field_id = null, $ref_string = null, $field_name = null,$base_form = null,$form_name=null) { 

     if(!$base_form) { 
      $base_form = $form_id; 
     } 
     App::import('model', 'Section'); 
     $zmr_sections = new Section(); 


     $sectionList = $zmr_sections->getSectionList($form_id); 


     if ($sectionList) { 

      $new_base_fields = $this->Field->find('list', array('joins' => array(
       array(
        'table' => 'field_types', 
        'alias' => 'Type', 
        'type' => 'left', 
        'foreignKey' => false, 
        'conditions' => array('Type.id = Field.field_type_id') 
       )), 'conditions' => array('Type.base_type != "formatting" AND Type.base_type != "reference" AND Type.base_type != "reverse reference"', 'Field.active=1', 'Field.section_id IN (' . $sectionList . ')'))); 


      if ($ref_field_id) { 
       $newFields = array(); 


       foreach ($new_base_fields AS $key => $field) { 
        if ($field_name) { 

         $field = $form_name . ' - ' . $field; 
        } 

        $newFields[$ref_string . '-' . $key] = $field; 

       } 
       $new_base_fields = $newFields; 
      } 
      if (!$base_fields) { 
       $base_fields = array(); 
      } 

      $base_fields = $base_fields + $new_base_fields; 


      $reference_fields = $this->Field->find('all', array('joins' => array(
       array(
        'table' => 'field_types', 
        'alias' => 'Type', 
        'type' => 'left', 
        'foreignKey' => false, 
        'conditions' => array('Type.id = Field.field_type_id') 
       )), 'conditions' => array('Type.base_type' => 'reference', 'Field.active=1', 'Field.section_id IN (' . $sectionList . ')'))); 


      foreach ($reference_fields AS $reference_field) { 

       $field_name = $reference_field['Field']['name']; 
       $query = "SELECT form_id FROM zmr_lists WHERE id=" . $reference_field['Field']['zmr_list_id']; 

       $list_data = $this->query($query); 
       $new_form_id = $list_data[0]['zmr_lists']['form_id']; 



       if($form_id == $base_form){ 
        $ref_string = $reference_field['Field']['id']; 
        $form_name = $this->Field->Section->Form->field('Form.label',array('Form.id'=>$new_form_id)); 

       }else{ 
        $ref_string .= '-'. $reference_field['Field']['id']; 
        $form_name .= '-'. $this->Field->Section->Form->field('Form.label',array('Form.id'=>$new_form_id)); 

       } 
       $base_fields = $this->get_reference_fields($new_form_id, $base_fields, $reference_field['Field']['id'], $ref_string, $field_name,$base_form,$form_name); 

      } 


     } 

     return $base_fields; 

    } 

================================= =
而且，對於數據集的樣本結果是：

[49-7-6-48] => Doctor - Primary Role 
[49-7-6-43] => Doctor - State 
[49-7-6-11] => Doctor - Password 
[49-7-6-2] => Doctor - Last Name 
[49-7-6-10] => Doctor - Username 
[49-7-6-28] => Doctor - City 
[49-7-6-1] => Doctor - First Name 
[49-7-6-24] => Doctor - Middle Initial 
[49-7-6-27] => Doctor - Address2 
[49-7-6-25] => Doctor - Date of Birth 
[49-7-6-47] => Doctor - Phone Number 
[49-7-6-26] => Doctor - Address1 

[49-7-5-68] => PSG Study - Patient ID Number 
[49-7-5-67] => PSG Study - Insurance Provider's Name 
[49-7-5-74] => PSG Study - PCP referral required? 
[49-7-5-73] => PSG Study - Pre-Authorization Obtained? 
[49-7-5-22] => PSG Study - Special Needs 
[49-7-5-76] => PSG Study - PCP Phone Number 
[49-7-5-69] => PSG Study - Group Number 
[49-7-5-70] => PSG Study - Relationship to Insured Member 
[49-7-5-71] => PSG Study - Referring Doctor 
[49-7-5-75] => PSG Study - If Yes, PCP Name: 
[49-7-5-72] => PSG Study - Diagnoses: 
[49-7-5-4-48] => Patient - Primary Role 
[49-7-5-4-43] => Patient - State 
[49-7-5-4-11] => Patient - Password 
[49-7-5-4-2] => Patient - Last Name 
[49-7-5-4-10] => Patient - Username 
[49-7-5-4-28] => Patient - City 
[49-7-5-4-1] => Patient - First Name 
[49-7-5-4-24] => Patient - Middle Initial 
[49-7-5-4-27] => Patient - Address2 
[49-7-5-4-25] => Patient - Date of Birth 
[49-7-5-4-47] => Patient - Phone Number 
[49-7-5-4-26] => Patient - Address1 

========================= 

誰能幫我找出一種算法正常地生成參考鏈？請注意，表單在技術上可能是指自己。例如，醫生可以有另一位醫生的參考（見習生或主管等）

Answer 1

我找到了答案。我知道ref_string的初始化是問題。我無法理解它。我加入這個遞歸函數調用後立即

if ($ref_string) { 
         $ref_array = explode('-', $ref_string); 
         array_pop($ref_array); 
         $ref_string = implode('-', $ref_array); 
         $form_name_array = explode('-', $form_name); 
         array_pop($form_name_array); 
         $form_name = implode('-', $form_name_array); 
        }