Since you are using MySQL 5.7, you don't have support for the JSON_TABLE() function, which would be the solution for the query you describe.
SELECT j.tag, COUNT(*) AS count
FROM mytable CROSS JOIN
JSON_TABLE(tags, '$.en[*]' COLUMNS(tag VARCHAR(10) PATH '$.') AS j
GROUP BY j.tag
ORDER BY count DESC;
I would strongly recommend that you store your multi-valued data in normal columns and rows, not in JSON. Then the query would become more straightforward:
SELECT `tag`, COUNT(*) AS count
FROM en_tags
GROUP BY `tag`
ORDER BY count DESC;
Answer from Bill Karwin on Stack OverflowExtract data from one special object out of JSON array in mySQL - Databases - SitePoint Forums | Web Development & Design Community
mysql - Extract JSON object inside of an JSON array then search base on ID - Database Administrators Stack Exchange
How to get a specific object in an JSON array in MySQL? - Stack Overflow
MySQL search json value by key in array - Stack Overflow
Videos
Yes it is possible to do it using the JSON_EXTRACT and JSON_SEARCH functions.
Let's say your table is named tbl_Jobs and the jobs column is of type JSON.
SELECT * FROM tbl_Jobs
WHERE JSON_EXTRACT(jobs, "$[*].target") = JSON_EXTRACT(jobs, "$[*].done")
AND JSON_EXTRACT(jobs, "$[*].status") != 0
AND JSON_SEARCH(jobs, 'one', DATE_SUB(CURDATE(), INTERVAL 1 DAY), NULL, "$[*].lastAction") IS NOT NULL
SELECT JSON_PRETTY(JSON_EXTRACT(jobs.jobs, CONCAT('$[', j.rownum-1, ']'))) AS object
FROM jobs
CROSS JOIN JSON_TABLE(
jobs.jobs, '$[*]' COLUMNS(
rownum for ordinality,
done int path '$.done',
target int path '$.target',
status int path '$.status'
)
) as j
WHERE j.target > j.done AND j.status != 0;
You also mentioned a condition on lastAction, but the example values you gave are not valid dates, so I'll leave that enhancement to you. The example above demonstrates the technique.
There is a way to solve this in SQL 5.7. I will go step by step in composing the solution. The goal is to find the strength of the knight.
I am going to use the same sample table as previous post:
create table mytable ( mycol json );
insert into mytable set mycol = '[{"Race": "Orc", "strength": 14}, {"Race": "Knight", "strength": 7}]';
First, get an array of only the races.
select json_extract(mycol, '$[*].Race') from mytable;
+----------------------------------+
| json_extract(mycol, '$[*].Race') |
+----------------------------------+
| ["Orc", "Knight"] |
+----------------------------------+
Then, search for the Knight in this array (and unquote it).
select json_unquote(json_search(json_extract(mycol, '$[*].Race'), 'one', 'Knight')) from mytable;
+------------------------------------------------------------------------------+
| json_unquote(json_search(json_extract(mycol, '$[*].Race'), 'one', 'Knight')) |
+------------------------------------------------------------------------------+
| $[1] |
+------------------------------------------------------------------------------+
Having found the index, get this element from the array.
select json_extract(mycol, json_unquote(json_search(json_extract(mycol, '$[*].Race'), 'one', 'Knight'))) from mytable;
+---------------------------------------------------------------------------------------------------+
| json_extract(mycol, json_unquote(json_search(json_extract(mycol, '$[*].Race'), 'one', 'Knight'))) |
+---------------------------------------------------------------------------------------------------+
| {"Race": "Knight", "strength": 7} |
+---------------------------------------------------------------------------------------------------+
Then get the strength of this element.
select json_extract(json_extract(mycol, json_unquote(json_search(json_extract(mycol, '$[*].Race'), 'one', 'Knight'))), '$.strength') as strength from mytable;
+----------+
| strength |
+----------+
| 7 |
+----------+
You can repeat this on other fields to create other columns.
You're essentially meaning to apply selection and projection to the array elements and object fields of your JSON document. You need to do something like a WHERE clause to select a "row" within the array, and then do something like picking one of the fields (not the one you used in your selection criteria).
These are done in SQL using the WHERE clause and the SELECT-list of columns, but doing the same with JSON isn't something you can do easily with functions like JSON_SEARCH() and JSON_CONTAINS().
The solution MySQL 8.0 provides is the JSON_TABLE() function to turn a JSON document into a virtual derived table — as though you had defined conventional rows and columns. It works if the JSON is in the format you describe, an array of objects.
Here's a demo I did by inserting your example data into a table:
create table mytable ( mycol json );
insert into mytable set mycol = '[{"Race": "Orc", "strength": 14}, {"Race": "Knight", "strength": 7}]';
SELECT j.* FROM mytable, JSON_TABLE(mycol,
'$[*]' COLUMNS (
race VARCHAR(10) PATH '$.Race',
strength INT PATH '$.strength'
)
) AS j;
+--------+----------+
| race | strength |
+--------+----------+
| Orc | 14 |
| Knight | 7 |
+--------+----------+
Now you can do things you normally do with SELECT queries, like selection and projection:
SELECT j.strength FROM mytable, JSON_TABLE(mycol, '$[*]'
COLUMNS (
race VARCHAR(10) PATH '$.Race',
strength INT PATH '$.strength'
)
) AS j
WHERE j.race = 'Orc'
+----------+
| strength |
+----------+
| 14 |
+----------+
This has a couple of problems:
You need to do this every time you query the JSON data, or else create a VIEW to do it.
You said you don't know the attribute fields, but to write a JSON_TABLE() query, you must specify the attributes you want to search and project in your query. You can't use this for totally undefined data.
I've answered quite a number of similar questions about using JSON in MySQL. I've observed that when you want to do this sort of thing, treating a JSON document like a table so you can apply condition in the WHERE clause to fields within your JSON data, then all your queries get a lot more difficult. Then you start feeling like you would have been better off spending a few minutes to define your attributes so you could write simpler queries.

