MySQL convert multiple columns into JSON - mysql

Question

I'm trying to convert multiple columns from one table into single JSON in another table in mysql database (version 5.7.16). I want use SQL query.
First table look like this
CREATE TABLE `log_old` (
`id` INT(10) UNSIGNED NOT NULL AUTO_INCREMENT,
`temperature` DECIMAL(5,2) NULL DEFAULT NULL,
`heating_requested` BIT(1) NULL DEFAULT NULL,
PRIMARY KEY (`id`),
)COLLATE='utf8_general_ci'
ENGINE=InnoDB;
Second table look like this
CREATE TABLE `log_new` (
`id` INT(10) UNSIGNED NOT NULL AUTO_INCREMENT,
'data' JSON,
PRIMARY KEY (`id`),
)COLLATE='utf8_general_ci'
ENGINE=InnoDB;
data JSON has same format in all rows of log_new table, it should look like this
{
temperature: value,
heatingRequested: false
}
for example log_old look like this
+--+-----------+-----------------+
|id|temperature|heating_requested|
+--+-----------+-----------------+
|1 | 12 | true |
+--+-----------+-----------------+
|2 | 14 | true |
+--+-----------+-----------------+
|3 | 20 | false |
+--+-----------+-----------------+
and I want to log_new looked like this
+--+-----------------------------------------+
|id| data |
+--+-----------------------------------------+
|1 |{temperature:12, heatingRequested: true} |
+--+-----------------------------------------+
|2 |{temperature:14, heatingRequested: true} |
+--+-----------------------------------------+
|3 |{temperature:20, heatingRequested: false}|
+--+-----------------------------------------+
I tried to use JSON_INSERT()
SELECT JSON_INSERT((SELECT data FROM log_new ), '$.temperature',
(SELECT temperature FROM log_old));
but this throw error "subquery returns more than 1 row"
I came with only solutions thats work to use while and do it row by row but this can take long time
DELIMITER //
CREATE PROCEDURE doLog()
BEGIN
SELECT COUNT(*) into #length from log_zone;
SET #selectedid = 1;
WHILE #selectedid < #length DO
SELECT temperature,heating_requested INTO #temperature,#heating_requested FROM log_old where id=#selectedid;
SELECT JSON_OBJECT('temperature',#temperature,'heatingRequested',#heating_requested) into #data_json;
SET #selectedid = #selectedid + 1;
INSERT INTO log_new (data) VALUES (#data_json);
END WHILE;
END;
//
CALL doLog()

Answer 1

As all your data are available on single lines, you don't need to use subqueries or loops to build the json object.
You can try something like :
INSERT INTO log_new (data)
SELECT json_object('temperature',log_old.temperature,'heatingRequested',log_old.heating_requested)
FROM log_old

Answer 2

Use a programming language or BI tool. Your question is very thought out but what I am missing is why does this have to be in mysql?
An RDMS, although many have reporting addons, is not intended to provide this low level manipulation. You are entering in a reporting realm and may need to focus on a view of your data outside of the database. You would be best served using Node, PHP, Python, and just about any actual programming language with strong mysql support (which is just about every modern language out there). BI tools include several free options like Pentaho/Kettle and Google's Data Studio and countless commercial BI options like Tableau and the like.
It is my strong belief that stored procedures, although they have a place, should not be responsible for application logic.