Answers for "SQL JSON_EXTRACT"

SQL JSON_EXTRACT

select JSON_UNQUOTE(JSON_EXTRACT(base, '$.scope')) as scope from t_name

Posted by: Guest on October-04-2021

mysql JSON_SEARCH LIKE

$query=$query->whereRaw('UPPER(education->"$[*].profession") LIKE UPPER("%' . $profession . '%")');

Posted by: Guest on December-08-2020

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