In the previous two articles, I have shared basic detail about MySQL 5.7 JSON Data Type.
Our next question is, How we can create an Index on JSON Data Type in MySQL 5.7.

As per the official document, actually we cannot directly create an Index on JSON column in MySQL.

We have to add generated column in Table by extracting require JSON Key data.
Once require JSON key data extracted, we can apply index on that column like any other column.

Actually, this is not a good way, but as of now we have only this option to create an indirect Index on JSON Column in MySQL 5.7.

Create sample table using JSON data type:

CREATE TABLE tbl_TestJSON
(
	ID INTEGER PRIMARY KEY
	,DepartName VARCHAR(250)
	,EmployeeDetails JSON
	,Address JSON
);

Insert few sample JSON formatted record:

INSERT INTO tbl_TestJSON
VALUES 
(
	1
	,'Sales'
	,'{"firstName": "Anvesh", "lastName": "Patel"}'
	,'
	{"address" : 
		{ 
			"India": "Hyderabad"
			,"USA": "Newyork"
		}
	}'
)
,(
	2
	,'Production'
	,'{"firstName": "Neevan", "lastName": "Patel"}'
	,'
	{"address" : 
		{ 
			"India": "Ahmedabad"
			,"USA": "Washington DC"
		}
	}'
)
,(
	3
	,'Animation'
	,'{"firstName": "Eric", "lastName": "Lorn"}'
	,'
	{"address" : 
		{ 
			"India": "Mumbai"
			,"USA": "Chicago"
		}
	}'
);

Create aditional generated column for JSON Field EmployeeDetails->firstName:

ALTER TABLE tbl_TestJSON ADD VirtualColumn VARCHAR(50) 
GENERATED ALWAYS AS (JSON_EXTRACT(EmployeeDetails, '$.firstName')) VIRTUAL;

Create Index on generated column which indirectly work as Index on JSON EmployeeDetails->firstName:

CREATE INDEX idx_tbl_TestJSON_VirtualColumn ON tbl_TestJSON(VirtualColumn);

Check the execution plan of below query, where you can find usage of newly created Index:

EXPLAIN
SELECT JSON_EXTRACT(EmployeeDetails, "$.firstName") 
FROM tbl_TestJSON 
WHERE VirtualColumn = 'Anvesh';