Select query from json format data

Question

I just want a select query where it can Select all user uid who have born after 1986 Year.

I tried this query but it only selects the matching number LIKE search.

SELECT uid FROM {profile_values} WHERE  value LIKE  '%1986%'

Here is the SQL query

CREATE TABLE IF NOT EXISTS `profile_values` (
  `fid` int(10) unsigned NOT NULL DEFAULT '0',
  `uid` int(10) unsigned NOT NULL DEFAULT '0',
  `value` text,
  PRIMARY KEY (`uid`,`fid`),
  KEY `fid` (`fid`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;

INSERT INTO `profile_values` (`fid`, `uid`, `value`) VALUES
(5, 1, 'a:3:{s:3:"day";s:2:"27";s:5:"month";s:1:"6";s:4:"year";s:4:"1986";}'),
(5, 3, 'a:3:{s:3:"day";s:2:"15";s:5:"month";s:1:"9";s:4:"year";s:4:"1910";}'),
(5, 4, 'a:3:{s:3:"day";s:2:"26";s:5:"month";s:1:"6";s:4:"year";s:4:"1986";}'),
(5, 5, 'a:3:{s:5:"month";s:1:"4";s:3:"day";s:2:"26";s:4:"year";s:4:"2014";}'),
(5, 6, 'a:3:{s:3:"day";s:2:"26";s:5:"month";s:1:"4";s:4:"year";s:4:"2014";}'),
(5, 7, 'a:3:{s:5:"month";s:1:"4";s:3:"day";s:2:"26";s:4:"year";s:4:"2014";}'),
(5, 8, 'a:3:{s:3:"day";s:2:"26";s:5:"month";s:1:"4";s:4:"year";s:4:"1987";}'),
(5, 17, 'N;'),
(5, 18, 'N;'),

Answer 1

You can try REGEXP. The REGEXP matching condition is similar to the LIKE but LIKE only performs simple pattern matching using the wildcards "%" and "_", REGEXP performs complex regular expression pattern matching allowing it to match a much greater range of string patterns than LIKE.

SELECT uid FROM {profile_values} 
WHERE value REGEXP '[1][9][8][6-9]|[1][9][9][0-9]|[2-9][0-9][0-9][0-9]';

Well here we make sure that our expression considers only 4 digit number and moreover number should be greater than 1986

so [1][9][8][6-9] -->> makes sure number is between 1986 to 1989
next [1][9][9][0-9] -->> makes sure number is between 1990 to 1999
last [2-9][0-9][0-9][0-9] -->> makes sure number is between 2000 to 9999

please refer http://www.tutorialspoint.com/mysql/mysql-regexps.htm for more