MySQL JSON: finding value of sibling element in sub-array - mysql

Question

I have the following (pseudo)JSON in a type JSON (LONGTEXT) column in my MariaDB 10.2
{"order":
{"otherstuff":...},
{"dates":
[
{
"typeId":2,
"date":"2019-05-21 09:00:00"
},
{
"typeId":4,
"date":"2019-05-21 10:00:00"
}
]
}
}
What I need is the order's date while I know which type I need (4).
An order can have a number of dates identified by their typeId. typeId 4 is not always in second position.
SELECT JSON_UNQUOTE(JSON_SEARCH(`json`, 'one', 4, NULL, '$.dates[*].typeId'))
// gives me: $.dates[1].typeId
My first thought now was to REPLACE typeId with date, but that complains about mixed collations.
How would I (more elegantly) reference the 'date' value here?
Also, the query is supposed to be the expression of a GENERATED column in my table. Since date id4 is not necessarily there for every order, I tried this:
SELECT IF(4 IN (JSON_EXTRACT(json, '$.dates[*].typeId')), 'yes', 'no')
// above condition evaluates to [2, 4]
I have trimmed away '[' and ']' but then it only gives me a 'yes' if 4 is first in the array (is it an array?).
So (without brackets):
[4, 7] -> yes
[2, 4] -> no
I'm assuming this doesn't get recognized as an array of values but a string. Then why does it give me 'yes' if my needle is in first position?
Instead of yes and no I obviously want to use the date and NULL.
The MySQL JSON functions are quite new to me. So maybe someone could point me in the right direction?

Answer 1

Try:
Option 1:
SELECT
JSON_UNQUOTE(
JSON_EXTRACT(
`json`,
REPLACE(
JSON_UNQUOTE(
JSON_SEARCH(
`json`,
'one',
4,
NULL,
'$.order.dates[*].typeId'
)
),
'typeId',
'date'
)
)
) `date`;
Option 2:
SELECT
IF(
JSON_CONTAINS(
JSON_EXTRACT(
`json`,
'$.order.dates[*].typeId'
),
4
),
'yes',
'no'
) `exists`;
See dbfiddle.