mysql 操作json

 

SELECT `id`,
`array1`

FROM
`mimeng_test`.`t_array`
LIMIT 0, 1000;

INSERT INTO `mimeng_test`.`t_array`(id,array1) VALUES(2,'{"result":0,"white":[0,1,2,3}')

 

1.修改json

UPDATE `mimeng_test`.`t_array` SET array1 = JSON_SET(array1,'$.result',1,'$.white',JSON_ARRAY(7,8,9))
WHERE id = 2

 

2.追加数组的值

UPDATE `mimeng_test`.`t_array` SET array1=JSON_ARRAY_APPEND(array1,'$.white',666) WHERE id = 2;

posted @ 2022-08-10 16:19  密蒙  阅读(228)  评论(0编辑  收藏  举报