mysql 使用JSON_EXTRACT() 取json值

mysql 取json字符串字段下的某个键的值

要求：mysql版本5.7及以上

SELECT JSON_EXTRACT('{"uid":"asas02234"}', "$.uid");

由于json的键值是带双引号。所以需要去掉双引号。

1.使用replace()做替换

select 
    replace(JSON_EXTRACT(infoJson,'$.uid'),'"','') uid
from users

laravel 里使用JSON_EXTRACT

$applyList = DB::table('invoice')
            ->select('applied_at','total_amount','invoice_form','invoice_file','reject_reason')
            ->selectRAW('replace(JSON_EXTRACT(invoice_detail,"$.owner_name"),\'"\',\'\') as owner_name')
            ->where('uid',Auth::id())->get();

2.使用 JSON_UNQUOTE()

SELECT JSON_UNQUOTE(JSON_EXTRACT('{"id":"3"}', "$.id"));

$applyList = DB::table('invoice')
            ->select('applied_at','total_amount','invoice_form','invoice_file','reject_reason')
            ->selectRAW('JSON_UNQUOTE(JSON_EXTRACT(invoice_detail,"$.owner_name")) as owner_name')
            ->where('uid',Auth::id())->get();