SQL replace all specified keys - mysql

Question

I have one column(varchar) containing only json string within one table. I want replace all keys with "" on that column. How can I do that using sql? My database is MySQL.
For example:
|--------------------------------------------------------------------|
| t_column |
|--------------------------------------------------------------------|
| {"name":"mike","email":"xxx#example.com","isManage":false,"age":22}|
|--------------------------------------------------------------------|
SELECT replace(t_column, regexp, "") FROM t_table
I expect:
mikexxx#example.comfalse22
How to write that regexp?

Answer 1

Start from
select t_column->'$.*' from test
This will return a JSON array of attribute values:
[22, "mike", "xxx#example.com", false]
This might be already all you need, and you can try something like
select *
from test
where t_column->'$.*' like '%mike%';
Unfortunately there seems to be no native way to join array values to a single string like JSON_ARRAY_CONCAT(). In MySQL 8.0 you can try REGEXP_REPLACE() and strip all JSON characters:
select regexp_replace(t_column->'$.*', '[" ,\\[\\]]', '') from test
which will return '22mikexxx#example.comfalse'.
If the values can contain one of those characters, they will also be removed.
Note: That isn't very reliable. But it's all I can do in a "simple" way.
See demo on db-fiddle.

Answer 2

I could be making it too simplistic, but this is just a mockup based on your comment. I can formalize it into a query if it fits your requirement.
Let's say you get your JSON string to this format where you replace all the double quotes and curly brackets and then add a comma at the end. After playing with replace and concat_ws, you are now left with:
name:mike,email:xxx#example.com,isManage:false,age:22,
With this format, every value is now preceded by a semicolon and followed by a comma, which is not true for the key. Let's say you now want to see if this JSON string has the value "mike" in it. This, you could achieve using
select * from your_table where json_col like '%:mike,%';

Answer 3

If you really want to solve the problem with your approach then the question becomes
What is the regex that selects all the undesired text from the string {"name":"mike","email":"xxx#example.com","isManage":false,"age":22} ?
Then the answer would be: {\"name\":\"|\"email\":\"|\",\"isManage\":|,\"age\":|}
But as others let you notice I would actually approach the problem parsing JSONs. Look up for functions json_value and json_query
Hope I helped
PS: Keep close attention on how I structured the bolded sentence. Any difference changes the problem.
EDIT:
If you want a more generic expression, something like select all the text that is not a value on a json-formatted string, you can use this one:
{|",|"\w+\":|"|,|}