我該如何推送這個2-D數組？ （PHP）

Question

在這個codeigniter模型中，我試圖推送每個產品的類別名稱。 我想看看數組一樣：

Array 
(
    [0] => Array 
     (
     [id] => 1 
     [name] => Game 1! 
     [category_id] => 3 
     [category] => games //the category element is in the [0] array. 
    ) 

    [1] => Array 
     (
      [id] => 2 
      [name] => Game 2 
      [category_id] => 3 
      [category] => games 
     ) 

這裏的陣列的外觀現在：

Array 
(
    [0] => Array 
     (
      [category] => games //i want this to be in the [1] array 
     ) 

    [1] => Array 
     (
     [id] => 1 
     [name] => Game 1! 
     [category_id] => 3 
    ) 

[2] => Array 
    (
     [category] => games 
    ) 

[3] => Array 
    (
     [id] => 2 
     [name] => Game 2 
     [category_id] => 3 
    ) 

這裏是我的功能得到所有產品，並把它們在數組中。

function getAllProducts(){ 
    $data=array(); 
    $Q=$this->db->get('products'); 
    if($Q->num_rows() > 0){ 
     foreach($Q->result_array() as $row){ 
      $Q2=$this->db->query("select name FROM categories 
      where id={$row['category_id']}"); 
      if($Q2->num_rows() > 0){ 
       foreach($Q2->result_array() as $row2){ 
        //Trouble Here: dont know how to push this into the array. 
        //is there a function that i can put in the $data[___] 
        //area so that it knows it is in the [0] element, then [1],etc? 
        $data[]['category']=$row2['name']; 
       } 

      }  
      $data[]=$row; 
     } 
    } 
    $Q->free_result(); 
    return $data; 
} 

Answer 1

要麼是：

foreach($Q->result_array() as $key=>$row){ 
     $data[$key]=$row; 
     $Q2=$this->db->query("select name FROM categories 
     where id={$row['category_id']}"); 
     if($Q2->num_rows() > 0){ 
      foreach($Q2->result_array() as $row2){ 
       $data[$key]['category']=$row2['name']; 
      } 

     }  

    } 

甚至更​​好，做出正確的查詢：

function getAllProducts(){ 
    $data=array(); 
    $Q=$this->db->query("SELECT p.*, c.name as category FROM products p, LEFT JOIN categories c ON c.id=p.category_id"); 
    if($Q->num_rows() > 0){ 
     foreach($Q->result_array() as $row){ 
      $data[] = $row; 
     } 
    } 
    $Q->free_result(); // <- needed here? don't know 
    return $data; 
} 

我不知道笨，所以我不知道這是否是完全正確的。但總的來說這要好得多，因爲你減少的數據庫訪問數量是總是好。如果您有n產品，請聯繫我們。

P.S .:我打賭CodeIgniter也提供了一些API來製作JOIN。閱讀documentation。

我剛纔看到，你可以這樣做：

$this->db->select('*'); 
$this->db->from('products'); 
$this->db->join('categories', 'categories.id = products.category_id', 'left'); 

$Q = $this->db->get(); 

利用API的。

Answer 2

我會嘗試一種不同的方法來減少您需要進行的總查詢次數。

$productRows = getAllProducts(); 
$categoryRows = getAllCategories(); 

$categoriesById = array(); 

foreach ($categoryRows as $categoryRow) { 

    $categoriesById[$categoryRow['id']] = $categoryRow; 
} 

$productsWithCategory = array(); 

foreach ($productRows as $productRow) { 

    $categoryId = $productRow['category_id']; 
    $productsWithCategory[] = array(
       'id' => $productRow['id'], 
       'name' => $productRow['name'], 
     'category_id' => $categoryId, 
      'category' => $categoriesById[$categoryId]['name'] 
    ); 
}