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Работа с JSON данными в СУБД Mysql 5.7
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Выборка из масива Тип поля в БД должен быть json | |
В поле хранится по типу [one,two,three,four,five] | |
SELECT * FROM `table_name` WHERE JSON_CONTAINS(`field`,'[value]'); | |
В поле хранится по типу ID ТАБЛИЦЫ 1 - ["one","two","three","four","five"] | |
SELECT * FROM `table_name` WHERE JSON_CONTAINS(`field`,'['\"value\"']'); | |
SELECT * FROM `table_name` WHERE JSON_CONTAINS(`field`,JSON_ARRAY('value')); | |
SELECT JSON_EXTRACT((SELECT `field` FROM `table_name` WHERE `id` = 1),'$[index]') as title; | |
Выборка из обьекта | |
В поле хранится по типу ID ТАБЛИЦЫ 2 - {"one":1,"two":2,"three":3,"four":4,"five":5} | |
SELECT * FROM `table_name` WHERE JSON_CONTAINS(`field`,'{\"key\":value}'); | |
SELECT * FROM `table_name` WHERE JSON_CONTAINS(`field`,JSON_OBJECT('value')); | |
SELECT * FROM `table_name` WHERE JSON_CONTAINS(`field`,JSON_OBJECT('key','value','key2','value2')); | |
SELECT JSON_EXTRACT((SELECT `field` FROM `table_name` WHERE `id` = 2),'$.*') as title; | |
SELECT JSON_EXTRACT((SELECT `field` FROM `table_name` WHERE `id` = 2),'$.key') as title; | |
Более сложная выборка | |
В поле хранится по типу ID ТАБЛИЦЫ 3 | |
{"0":"one", | |
"1":{"one":1,"two":2,"five":["value1","value2"],"four":4,"three":3}, | |
"2":["six","seven","eight"],"3":"five","key":"value"} | |
SELECT JSON_EXTRACT((SELECT `field` FROM `table_name` WHERE `id` = 3),'$.*.key') as title; | |
SELECT JSON_EXTRACT((SELECT `field` FROM `table_name` WHERE `id` = 3),'$.*.key[index]') as title; | |
SELECT `field`, `field2`->'$.*.key' AS `field2` FROM `table_name`; | |
SELECT `field`, `field2` AS `field2` FROM `table_name` WHERE `field2`->'$.*.key[index]' IS NOT NULL; | |
Источник | |
https://webformyself.com/rabota-s-json-dannymi-v-subd-mysql-5-7/ | |
https://www.youtube.com/watch?time_continue=1082&v=riOQT4HS7yI |
Да вот только вся проблема, что с русским текстом в json это все не работает
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Блин спасибо! Долго гуглил как массив сделать с динамическими данными и вот оно из примера - JSON_OBJECT !