从 JSON_OBJECT 按值排序 mysql 结果

Question

I have MariaDB 10.2 and this SQL:

SELECT products.*,
  CONCAT('[', GROUP_CONCAT(JSON_OBJECT(
  'id', V.id,
  'price', V.price
  ) ORDER BY V.price ASC), ']') AS variants,
FROM products
LEFT JOIN products_variants V ON V.products_id = products.id
GROUP BY products.id
LIMIT 0,10

Result is:

Array (
  [0] => Array (
    [id] => 1,
    [variants] => [{"id": 1, "price": 100},{"id": 2, "price": 110}]
  )
  [1] => Array (
    [id] => 2,
    [variants] => [{"id": 3, "price": 200},{"id": 4, "price": 210}]
  )
)

I need to sort the products according to the price of the first variation of each product.

• Product id 1 must be the first because the first variant price is 100
• Product id 2 must be the second because the first variant price is 200 and 200 > 100

I try:

ORDER BY JSON_EXTRACT(`variants`, '$[0].price)

but get error:

Reference 'variants' not supported (reference to group function)

Answer 1

I think you want:

order by min(v.price)

There is no need to parse the JSON object.

Answer 2

This is an example of when one should not hide a column ( price ) inside a column (in JSON, in this case). Instead, keep price as its own column to make it simple for MySQL to access for sorting (as you needed) or filtering.

It is OK to also keep the column inside the JSON string, so that the JSON is "complete".

The problem was that you had the wrong quotes around variant . Backtics (which you used) is for column and table names. You needed apostrophes or double-quotes ( ' or " ) because variant is just a string in this context.)