Mysql search in json array with Like operator - php

Question

I have a problem with MySQL json type. I want to search in mysql json array and get rows. My json data this
{
"id": 361,
"email": "example#outlook.com",
"code": null,
"username": null,
"permissions": null,
"created_at": "2019-03-01 16:09",
"updated_at": "2019-03-01 16:09",
"user_profile": {
"id": 361,
"name": "Jhon",
"surname": "Doe",
"father_name": "NED",
"birthday": "1994-12-15",
"fin_code": "6A56BS7",
"ssn": "AAA12830157",
"account_number": "account123",
"country": "USA",
"city": "NEW YORK",
"address": "EXample r",
"address_n": "Khani/Bina",
"mobile_phone": "(($717865643",
"phone": "0123456789",
"additional_email": "e.example#gmail.com",
"education": [
{
"endDate": "2020-06",
"startDate": "2015-09",
"profession": "Computer Since",
"university": "State University",
"educationType": 99
}
],
"language": [
{
"id": 102,
"level": 106
},
{
"id": 103,
"level": 106
},
{
"id": 104,
"level": 107
}
],
"computer_skills": [
{
"level": 106,
"skill": "php"
},
{
"level": 107,
"skill": "java"
}
],
"family_composition": [
{
"name": "Jhon",
"level": 126,
"surname": "Snow",
"birthday": "1992-02-08",
"fatherName": "Ned"
},
{
"name": "Jhon",
"level": 126,
"surname": "Snow",
"birthday": "1992-05-18",
"fatherName": "Ned"
}
],
"experience": [
{
"job": 128,
"time": 22,
"level": 8,
"salary": 2200,
"jobSign": 128,
"jobStatus": 267,
"startDate": "2012-12-12",
"orderNumber": "123dsa",
"jobSituation": 273,
"additionalSalary": 800
}
],
"reproach": [
{
"doc": 228,
"date": "2011-11-11",
"note": "Some reason",
"level": 225,
"number": "123dsa",
"reason": "islemir",
"article": 233
}
],
"additional_information": "All is work",
"citizen": {
"id": 5,
"name": "United States"
},
"cities": {
"id": 21,
"name": "New York"
},
"countries": {
"id": 89,
"name": "Unated States"
},
"gender": {
"id": 1,
"name": "Man"
},
"marital": {
"id": 4,
"name": "Single"
},
"work": {
"id": 269,
"name": "Able"
},
"party": {
"id": 10,
"name": "Digər"
},
"military": {
"id": 121,
"name": "OK"
},
"institution": null
}
}
I want to search like this:
WHERE education.'$[*].profession' Like %Computer%
But this syntax is not working. Thank you for replying. I developed my project in Laravel 5.7 if this is help for any suggestion. I don't use JSON_SEARCH() because this function returns the key but I need to return rows for my search query.

Answer 1

If you are on MySQL 5.7, this should work
.....WHERE JSON_EXTRACT(education , "$.profession") Like '%Computer%';

Answer 2

I found this solution and it worked for me:
UPPER(education->"$[*].profession") LIKE UPPER("% Computer %")
For Laravel syntax I write the PHP code like this:
$query=$query->whereRaw('UPPER(education->"$[*].profession") LIKE UPPER("%' . $profession . '%")');

Answer 3

Tested with MySQL 5.7:
SELECT 'found it!' FROM whatever_your_table_name_is
WHERE whatever_your_column_name_is->>'$.user_profile.education[*].profession'
LIKE '%Computer%';
Output, showing the WHERE clause matches the document:
+-----------+
| found it! |
+-----------+
| found it! |
+-----------+
As with most other questions about JSON in MySQL, I think you would be better off storing data in normalized tables.