2015-11-20 30 views
-1

我正在研究Codeigniter,我有以下表格,我希望首先打印一個主題名稱,然後打印出所有子主題,如下所示:無法從Codeigniter的foreach循環中的兩個表中獲取記錄

主題:

----------------------- 
id |  Subjects Name 
----------------------- 
1 |  Mathematics 
2 |  Physics 
3 |  Chemistry 

小節:

--------------------------------------------- 
id | subject_id | SubsectionsName 
--------------------------------------------- 
1  2    Electromagnetism 
2  1     Algebra 
3  3    Organic Chemistry 
4  1   Geometry and Topology 
5  2     Mechanics 

我想下面的輸出

Mathematics 
Algebra 
Geometry and Topology 
Physics 
Electromagnetism 
Mechanics 
Chemistry 
Organic Chemistry 

對我使用下面的代碼,但它不給我正確的輸出。

$condition = "id =" . "'" . $subject_id . "'"; 
$this->db->select('*'); 
$this->db->from('subject'); 
$this->db->where($condition); 
$query = $this->db->get(); 

if ($query->num_rows() > 0) { 

    foreach($query->result_array() as $row) 
    { 
     $rows[] = $row; 
     $condition = "subject_id =" . "'" . $row['id'] . "'"; 
     $this->db->select('*'); 
     $this->db->from('subsections'); 
     $this->db->where($condition); 
     $query_course_material = $this->db->get(); 
     foreach($query->result_array() as $row1) 
     { 
      $rows[] = $row1; 
     } 

    } 
    return $rows; 

} else { 
return false; 
} 

回答

1

foreach是具有不正確的變量。它應該有$query_course_material而不是$query。請檢查以下內容:

$query_course_material = $this->db->get(); 
foreach($query_course_material->result_array() as $row1) 
{ 
    $rows[] = $row1; 
} 
+0

@ user3653474:解決了您的問題? – AnkiiG

+0

是AG21感謝您的支持。 – user3653474

+0

但爲什麼-1這個問題。 – user3653474

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