如何从mysql多对多选择中将单个值组合为数组？

Question

我有三个表，一个用于商业列表，一个用于dining_types，一个用于dining_listing_types。

business_listings：

 +--------+------------------+------------------+
 |  id    |  business_type   |   business_name  |
 +--------+------------------+------------------+
 |  1     |       1          |   china house    |
 +--------+------------------+------------------+

dining_listing_types：

 +--------+------------------+------------------+
 |  id    |    listing_id    |    type_id       |
 +--------+------------------+------------------+
 |   1    |        1         |        1         |
 +--------+------------------+------------------+
 |   2    |        1         |        3         |
 +--------+------------------+------------------+

dining_types：

 +--------+------------------+
 |  id    |      type        |
 +--------+------------------+
 |  1     |   Asian          |
 +--------+------------------+
 |  2     |   Italian        |
 +--------+------------------+
 |  3     |   Chinese        |
 +--------+------------------+

如您所见，列表可以有多种类型，dining_listing_types表用作“中间人”表。

但是，当我尝试在查询中使用联接时，我会找回两条记录

SELECT bl.id AS id, bl.business_name AS name, dt.type AS type
FROM business_listings bl 
JOIN dining_listing_types dlt ON bl.id = dlt.listing_id 
JOIN dining_types dt ON dlt.type_id = dt.id

结果：

 array(2) { [0]=> object(stdClass)#7 (3) { ["id"]=> string(1) "1" ["name"]=> 
 string(23) "China House" ["type"]=> string(5) "Asian" } [1]=> 
 object(stdClass)#8 (3) { ["id"]=> string(1) "1" ["name"]=> string(23) "China 
 House" ["type"]=> string(18) "Chinese" } }

我不要那个。我尝试了一个GROUP BY bl.id，我找回了一条记录，但另一种类型丢失了：

SELECT bl.id AS id, bl.business_name AS name, dt.type AS type
FROM business_listings bl 
JOIN dining_listing_types dlt ON bl.id = dlt.listing_id 
JOIN dining_types dt ON dlt.type_id = dt.id
GROUP BY bl.id

结果：

 array(1) { [0]=> object(stdClass)#7 (3) { ["id"]=> string(1) "1" ["name"]=> 
 string(23) "China House" ["type"]=> string(5) "Asian" } }

我想要的结果：

array(1) { [0]=> object(stdClass)#7 (3) { ["id"]=> string(1) "1" ["name"]=> 
string(23) "China House" ["type"]=> array(2) ["Asian", "Chinese"] } }

Answer 1

正如@rickdenhaan评论的那样，GROUP_CONCAT()似乎非常接近你所寻找的。这不会生成types数组，而是以逗号分隔的列表。另一个奇特的选择是生成一个JSON数组，但我不确定这在这里真的有用。

为了使其按预期工作，您需要将business_name添加到非聚合列的列表中：

SELECT bl.id AS id, bl.business_name AS name, GROUP_CONCAT(dt.type) AS types
FROM business_listings bl 
INNER JOIN dining_listing_types dlt ON bl.id = dlt.listing_id 
INNER JOIN dining_types dt ON dlt.type_id = dt.id
GROUP BY bl.id, bl.business_name

注意：正如@spencer7593评论的那样，如果我们需要返回没有相关LEFT JOIN的INNER JOIN，那么使用business_listingss而不是dining_typess可能是一个好主意。

Answer 2

根据您的输出数据，我假设您正在使用PHP。如果你正在使用mysqli类（连接$conn），你可以用这样的东西实现你想要的东西（使用@rickdenhaan建议的GROUP_CONCAT）：

$sql = 'SELECT bl.id AS id, bl.business_name AS name, GROUP_CONCAT(dt.type) AS type
FROM business_listings bl 
LEFT JOIN dining_listing_types dlt ON bl.id = dlt.listing_id 
LEFT JOIN dining_types dt ON dlt.type_id = dt.id
GROUP BY bl.id, bl.business_name';
$result = $conn->query($sql) or die($conn->error);
while ($obj = $result->fetch_object()) {
    $obj->type = explode(',', $obj->type);
    // do something with $obj
}