我试图在PHP组查询结果作为一个有效的方式关联数组。我可以使用多个嵌套的foreach循环，检查当前id字段是否与每个数组组的前一个匹配，但我猜有更好/更有效的解决方案来解决这个问题。有人有聪明的解决办法吗?理想的通用函数或方法，可以组任何查询结果集给定的关键字段或嵌套的关键字段数组?
下面是查询结果数组:
Array
(
[0] => Array
(
[look_id] => 3
[look_name] => Test Look
[look_description] => description here
[clothing_article_id] => 1
[clothing_article_name] => Coat
[look_clothing_article_attribute_id] => 1
[look_clothing_article_attribute_name] => Purple
[clothing_brand_name] => Gap
[clothing_article_brand_price] => 40.00
[clothing_article_brand_attribute_price] => 50.00
[clothing_article_brand_attribute_name] => Purple
)
[1] => Array
(
[look_id] => 3
[look_name] => Test Look
[look_description] => description here
[clothing_article_id] => 1
[clothing_article_name] => Coat
[look_clothing_article_attribute_id] => 2
[look_clothing_article_attribute_name] => Black
[clothing_brand_name] => Gap
[clothing_article_brand_price] => 40.00
[clothing_article_brand_attribute_price] =>
[clothing_article_brand_attribute_name] =>
)
[2] => Array
(
[look_id] => 3
[look_name] => Test Look
[look_description] => description here
[clothing_article_id] => 2
[clothing_article_name] => Pants
[look_clothing_article_attribute_id] => 3
[look_clothing_article_attribute_name] => Cuffed
[clothing_brand_name] =>
[clothing_article_brand_price] =>
[clothing_article_brand_attribute_price] =>
[clothing_article_brand_attribute_name] =>
)
)
这就是我要转换成的数组:
Array
(
'looks' => Array(
[0] => Array(
'id' => 3,
'name' => 'Test Look',
'description' => 'description here',
'clothingArticles' => Array(
[0] => Array(
'id' => 1,
'name' => 'Coat',
'attributes' => Array(
[0] => Array(
'id' => 1,
'name' => 'Purple'
'brands' => Array(
[0] => Array(
'name' => 'Gap',
'price' => 50.00
)
)
),
[1] => Array(
'id' => 2,
'name' => 'Black'
)
),
'brands' => Array(
[0] => Array(
'name' => 'Gap',
'price' => 40.00
)
)
),
[1] => Array(
'id' => 2,
'name' => 'Pants',
'attributes' => Array(
[0] => Array(
'id' => 3,
'name' => 'Cuffed'
'brands' => Array()
)
),
'brands' => Array()
)
)
)
)
)
分组关系解释:
一个look可以有1个或多个服装物品。一件服装可以有一个或多个服装品牌。一篇服装文章可以有一个或多个属性。一个服装品牌可以有1个或多个属性。
这不应该太可怕;而不是对内部数组使用数字索引，您应该使用任何唯一标识符，以便您可以对其进行分组。这并不妨碍您在数组本身中使用该标识符。
//seems a little pointless?
$looks = array('looks' => array());
$looks = &$looks['looks'];
while ($row = $result->fetch()) {
if (!isset($looks[$row->look_id])) {
$looks[$row->look_id] = array(
'id' => $row->look_id,
'name' => $row->look_name,
'description' => $row->look_description,
'clothingArticles' => array()
);
}
$look = &$looks[$row->look_id];
if (!isset($look[$row->clothing_article_id])) {
//continue this process as needed
执行尽可能多的查询，每个表一个，所有查询都在相同的列上排序，并将行编织在一起:
$q1 = "SELECT * FROM look ORDER BY look_id";
$q2 = "SELECT look_id, article.* FROM look INNER JOIN article USING(look_id) ORDER BY look_id, article_id";
$q3 = "SELECT look_id, article_id, attribute.* FROM look INNER JOIN article USING(look_id) INNER JOIN attribute USING(article_id) ORDER BY look_id, article_id, attribute_id";
// loop on looks
// inner loop on article and add to look
// inner loop on attribute and add to article
或者你可以使用现成的ORM库，比如doctrine