在一個查詢

Question

例如MySQL的多重選擇我有幾個表：
產品：

| product_id | name | price | 
| 1   | apple | 20.32 | 
| 2   | pear | 9.99 | 
| 3   | banana | 1.5 | 

產品屬性：

| attr_id | name | value | 
| 1  | weight | 10 kg | 
| 2  | date | 2013 | 
| 3  | color | red | 

...等等。
最後產品屬性關係表：

| product_id | attr_id | 
| 1   | 3  | 
| 2   | 1  | 
| 1   | 2  | 
| 3   | 2  | 

我的問題：是否有可用的構建體，它返回一個選擇請求查詢產品1和2在下面的數據結構（或類似）？現在我應該首先運行deveral select請求「where product_id IN（1，2）」，然後循環選擇它們的屬性。
對不起，我英文不好：]

array(
    [0] = array(
      product_id = 1, 
      name = apple, 
      attributes= array(
         [0] => array(
          attr_id = 3, 
          name = color, 
          value = red, 
         ), 
         [0] => array(
          attr_id = 2, 
          name = date, 
          value = 2013, 
         ) 

        ), 
    ), 
    [1] = array(
      product_id = 2, 
      name = apple, 
      attributes= array(
         [0] => array(
          attr_id = 1, 
          name = veight, 
          value = 10 kg, 
         ), 
        ), 
    ) 
) 

Answer 1

這不僅是查詢的問題，也是PHP代碼。這將適合：

$rSelect = mysqli_query('SELECT 
    products.id AS record_id, 
    products.name AS products_name, 
    products.price AS product_price, 
    attributes.id AS attribute_id, 
    attributes.name AS attribute_name, 
    attributes.value AS attribute_value 
    FROM 
    products 
    LEFT JOIN products_attributes 
     ON products.id=products_attributes.product_id 
    LEFT JOIN attributes 
     ON products_attributes.attr_id=attributes.id', $rConnect); 

$rgResult = []; 
while($rgRow = mysqli_fetch_array($rSelect)) 
{ 
    $rgResult[$rgRow['record_id']]['product_id'] = $rgRow['record_id']; 
    $rgResult[$rgRow['record_id']]['name']   = $rgRow['product_name']; 
    $rgResult[$rgRow['record_id']]['price']  = $rgRow['product_price']; 
    $rgResult[$rgRow['record_id']]['attributes'][] = [ 
     'attr_id' => $rgRow['attribute_id'], 
     'name' => $rgRow['attribute_name'], 
     'value' => $rgRow['attribute_value'], 
    ]; 
}; 
//var_dump($rgResult); 

Answer 2

你正在尋找的查詢是：

select 
    p.product_id, 
    p.name as product_name, 
    a.attr_id, 
    a.name as attr_name, 
    a.value 
from 
    products as p 
    inner join `product-attribute` as pa on p.id = pa.product_id 
    inner join attribute as a on pa.attr_id = a.attr_id 

，將返回從那裏你可以創建一個多維數組的簡單結果表。

您可能會注意到在此查詢兩兩件事：

• 我使用反引號`周圍表名product-attribute，因爲它包含的名字，這是保留一個破折號-。所以你需要通過反引號來避開這個名字。
• 因爲字段名稱name不明確，我給他們一個別名（product_name/attr_name），以便他們以後仍然可以通過別名引用。