MySQL nested JSON column search and compare to with today's day - mysql

Question

I am using MySQL Version 5.7.28. I am having json data like below.
CREATE TABLE `week2` (
`id` int(11) NOT NULL AUTO_INCREMENT,
`type` smallint(1),
`json` text ,
PRIMARY KEY (`id`)
) ENGINE=InnoDB AUTO_INCREMENT=1 DEFAULT CHARSET=utf8;
INSERT INTO week2(id,type,json)
VALUES
(121,1,'[{"weekdays":"Sunday"},{"weekdays":"Monday"},{"weekdays":"Tuesday"},{"weekdays":"Wednesday"},{"weekdays":"Thursday"},{"weekdays":"Friday"},{"weekdays":"Saturday"}]'),
(122,1,'[{"weekdays":"Sunday"},{"weekdays":"Monday"}]'),
(123,2,'[{"start_time":"08:00 AM","end_time":"10:00 PM"}]');
As you see, the json column has nested JSON data. so here i am looking to compare today's day ( which is Saturday and we are currently in between start time and end time)
Expected Result:
(121,1,'[{"weekdays":"Sunday"},{"weekdays":"Monday"},{"weekdays":"Tuesday"},{"weekdays":"Wednesday"},{"weekdays":"Thursday"},{"weekdays":"Friday"},{"weekdays":"Saturday"}]'),
(123,2,'[{"start_time":"08:00 AM","end_time":"10:00 PM"}]');

Answer 1

You need to detect day name (1) and time period (2), and combine those two conditions by OR operator at the end.
For (1) : Detect the current day's name by using DAYNAME() function and search whether exists in the JSON data containing weekday keys through use of JSON_CONTAINS function.
For (2) : TIME() function and CASTing strings to TIME data type might be used with a trick to add 12 hours iterations for the cases of PM type times.
So, consider using :
SELECT *
FROM `week2`
WHERE JSON_CONTAINS(`json`->>'$[*].weekdays', CONCAT('"',DAYNAME( NOW() ),'"')) = 1
OR
(
TIME(ADDTIME(NOW(),"8:00:00")) >=
CASE WHEN INSTR(REPLACE(`json`->>'$[0].start_time',"12:00 AM","00:00 AM"),"PM")>0
THEN
CAST(CONCAT(MOD((TIME_FORMAT(REPLACE(`json`->>'$[0].start_time',"12:00 AM","00:00 AM"), "%T")+12),24),":00 AM")
AS TIME)
ELSE
CAST((REPLACE(`json`->>'$[0].start_time',"12:00 AM","00:00 AM")) AS TIME)
END
AND
TIME(ADDTIME(NOW(),"8:00:00")) <=
CASE WHEN INSTR(REPLACE(`json`->>'$[0].end_time',"12:00 AM","00:00 AM"),"PM")>0
THEN
CAST(CONCAT(MOD((TIME_FORMAT(REPLACE(`json`->>'$[0].end_time',"12:00 AM","00:00 AM"), "%T")+12),24),":00 AM")
AS TIME)
ELSE
CAST(REPLACE(`json`->>'$[0].end_time',"12:00 AM","00:00 AM") AS TIME)
END)
Demo

Answer 2

Perhaps, something like this:
SELECT *,CURRENT_TIME
FROM
(SELECT *,JSON_UNQUOTE(JSON_EXTRACT(`json`,'$[0].start_time')) AS st,
JSON_UNQUOTE(JSON_EXTRACT(`json`,'$[0].end_time')) AS et FROM week2) V
WHERE (JSON_SEARCH(`json`, 'one', DAYNAME(CURDATE())) IS NOT NULL
OR
CURRENT_TIME hour >=
CASE WHEN st LIKE '%AM%' THEN REPLACE(st,' AM',':00')
WHEN st LIKE '%PM%' THEN SEC_TO_TIME(TIME_TO_SEC(REPLACE(st,' PM',':00'))+43200)
END
AND
CURRENT_TIME hour <=
CASE WHEN et LIKE '%AM%' THEN REPLACE(et,' AM',':00')
WHEN et LIKE '%PM%' THEN SEC_TO_TIME(TIME_TO_SEC(REPLACE(et,' PM',':00'))+43200)
END);
First query is using JSON_EXTRACT to extract the time and JSON_UNQUOTE to remove the (") then make it as a sub-query.
On the outer-query, find the weekdays value using JSON_SEARCH compare with today's DAYNAME(CURDATE()); if JSON_SEARCH doesn't find it, it will return NULL, hence the IS NOT NULL condition is used.
Append OR to compare CURRENT_TIME with the time value extracted from the json field. But first, using CASE expression to find out if it's AM or PM; if it's AM, just REPLACE the AM with seconds hand (:00). If it's PM, do the REPLACE like AM then convert the time value to seconds using TIME_TO_SEC then add 43200 seconds (12 hours) and convert again to standard hour:minute:second format using SEC_TO_TIME.
P/S: It was my attempt to answer but I didn't post it because I get wrong result with the time.. Only when I put CURRENT_TIME in the fiddle that I realized that the fiddle time zone is different that mine. Also, I was testing on my local DB; which is MariaDB and the results of converting string to time using STR_TO_TIME is different. Honestly, in my opinion, if this is MariaDB, I think there's a chance that the query is much shorter.