PHP MySQL连接表作为JSON

参见英文答案 > How create json format with group-concat mysql?????????????????????????????????????6个
我有两个不同的表 – 一个称为“产品”,其中包含有关在线商店的产品信息,另一个称为“图像”,其中包含相应的产品图像(每个产品有几个图像).
我想在主表“products”上加入表“images”,并以JSON格式输出结果.

表“产品”如下所示(简化)：

id  | name
----|--------
57  | apple
58  | tomato
59  | ...

并且产品图像的表格“图像”：

img_id | img         | p_id |?listorder
-------|-------------|------|----------
32     | apple1.jpg  | 57   | 1
33     | apple2.jpg  | 57   | 2
34     | tomato1.jpg | 58   | 1
35     | ...         | ...  |?...

到现在为止,我的查询是这样的：

$sql = "SELECT
        p.id as p_id,p.name as p_name,i.*
        FROM products p
        JOIN (SELECT * FROM images WHERE listorder=1)
        AS i ON (i.p_id = p.id)";

因此输出(在将数据提取到数组并转换为JSON之后)看起来像这样：

[{
  "id": "57","name": "apple","img": "apple1.jpg"
},{
  "id": "58","name": "tomato","img": "tomato1.jpg"
}]

所以我的问题是：我如何输出以下内容？

[{
  "id": "57","img": [{"img_id": "32","img": "apple1.jpg"},{"img_id": "33","img": "apple2.jpg"}]
},"img": [{"img_id": "34","img": "tomato1.jpg"}]
}]

SELECT
    p.id as p_id,group_concat(concat('{"img_id":"',i.img_id,'"img":"',i.img,'"}') separator ',') as img
FROM products p
    JOIN (SELECT * FROM images WHERE listorder=1)
    AS i ON (i.p_id = p.id)
GROUP BY p.id,p.name