How to skip NULL values from MySQL CONCAT and GROUP_CONCAT - mysql

Question

I have this query that returns the name of the product and the array of images as a JSON string. However, when there is no image attached to a product, I would like this query to return an empty array for the images property.
Currently, it returns this when there is no product image found:
{ "name": "Product Name", "images": [{"id": null, "slug": null}]}
I tried to add an IF condition into the CONCAT method, but it returns the same response.
SELECT p.name,
CONCAT('[',
IF(i.id = NULL,
'',
GROUP_CONCAT(DISTINCT (
JSON_OBJECT(
'id', i.id,
'slug', i.slug
)
))
),
']') AS images
FROM products AS p
LEFT JOIN _product_images AS pi ON pi.pId = p.id
LEFT JOIN images AS i ON i.id = pi.iId
WHERE p.id = 4;
Thank you!

Answer 1

As other have mentioned, i.id = NULL will always evaluate to NULL. But your approach is needlessly complicated and would raise an error on a strictly configured server. On db-fiddle I get the following error:
ER_MIX_OF_GROUP_FUNC_AND_FIELDS: In aggregated query without GROUP BY,
expression #2 of SELECT list contains nonaggregated column
'test.i.id'; this is incompatible with sql_mode=only_full_group_by
demo
So the check i.id IS NULL needs to be done within the GROUP_CONCAT() function:
SELECT p.name,
CONCAT('[',
GROUP_CONCAT(DISTINCT (
IF (i.id IS NULL, '',
JSON_OBJECT(
'id', i.id,
'slug', i.slug
)
)
)),
']') AS images
FROM products AS p
LEFT JOIN _product_images AS pi ON pi.pId = p.id
LEFT JOIN images AS i ON i.id = pi.iId
WHERE p.id = 4
However - You can avoid the check, when you use an INNER JOIN. But the INNER JOIN will also ignore the data from the products table - So I would do that JOIN in a correlated subquery. An finally you can use JSON_ARRAYAGG() to generate a JSON array.
demo
SELECT p.name, COALESCE((
SELECT JSON_ARRAYAGG(JSON_OBJECT(
'id', i.id,
'slug', i.slug
))
FROM _product_images AS pi
JOIN images AS i ON i.id = pi.iId
WHERE pi.pId = p.id
), JSON_ARRAY()) AS images
FROM products AS p
#WHERE p.id = 4;
demo

Answer 2

You should be using IS NULL to check for a NULL value:
SELECT
p.name,
CONCAT('[',
GROUP_CONCAT(
IF(id IS NULL,
'',
DISTINCT JSON_OBJECT('id', i.id, 'slug', i.slug))),
']') AS images
FROM products AS p
LEFT JOIN _product_images AS pi ON pi.pId = p.id
LEFT JOIN images AS i ON i.id = pi.iId
WHERE p.id = 4
GROUP BY p.name;
As a side note, DISTINCT is not a function, and you should not be using it as such, so I removed the function parentheses which you were using.
Edit: Updated SQL to satisfy the group restrictions. The IF-Statement for checking i.id outside of the GROUP_CONCAT would fail, due to multiple image items per row.

Answer 3

IF(i.id = NULL, ...) returns NULL, for whatever value has i.id (even NULL), which is a falsy value.
You want to use instead IF (i.id IS NULL, ...)
From documentation :
You cannot use arithmetic comparison operators such as =, <, or <> to test for NULL. To demonstrate this for yourself, try the following query:
mysql> SELECT 1 = NULL, 1 <> NULL, 1 < NULL, 1 > NULL;
+----------+-----------+----------+----------+
| 1 = NULL | 1 <> NULL | 1 < NULL | 1 > NULL |
+----------+-----------+----------+----------+
| NULL | NULL | NULL | NULL |
+----------+-----------+----------+----------+