【求进】MySQL 5.7+如何Where查找JSON数组内对象的值

如，现在有张表 有 id、grade 两个字段，grade是JSON类型如果我要根据 id 值查找对应的JSON对象 如何查找grade列JSON格式示例如下
[
    {
        "id": "b814d6ac-68de-4f69-9e41-5b9de58ed65f",
        "name": "一天",
        "time": 1,
        "type": "day",
        "price": 55
    },
    {
        "id": "bbfc6bfe-2f81-40a5-b027-1e6e0b3bdeee",
        "name": "三天",
        "time": 3,
        "type": "day",
        "price": 70
    },
    {
        "id": "2a7fdfd6-8056-417b-8f26-f0f595265373",
        "name": "一周",
        "time": 7,
        "type": "day",
        "price": 300
    },
    {
        "id": "3b9a2ba9-5dad-40be-8e93-83f50e57396d",
        "name": "半小时",
        "time": 0.5,
        "type": "hour",
        "price": 7
    },
    {
        "id": "4facc993-3409-41eb-b078-8959b16707e0",
        "name": "一小时",
        "time": 1,
        "type": "hour",
        "price": 14
    },
    {
        "id": "77642501-5f1d-4757-b277-9df7571637cb",
        "name": "三小时",
        "time": 3,
        "type": "hour",
        "price": 30
    }
]
我现在只能查到 数组的所有内容SQL:
SELECT grade->'$[*]' FROM levels WHERE grade->'$[*].id' like '%77642501-5f1d-4757-b277-9df7571637cb%';结果：
[{"id": "b814d6ac-68de-4f69-9e41-5b9de58ed65f", "name": "一天", "time": 1, "type": "day", "price": 55}, {"id": "bbfc6bfe-2f81-40a5-b027-1e6e0b3bdeee", "name": "三天", "time": 3, "type": "day", "price": 70}, {"id": "2a7fdfd6-8056-417b-8f26-f0f595265373", "name": "一周", "time": 7, "type": "day", "price": 300}, {"id": "3b9a2ba9-5dad-40be-8e93-83f50e57396d", "name": "半小时", "time": 0.5, "type": "hour", "price": 7}, {"id": "4facc993-3409-41eb-b078-8959b16707e0", "name": "一小时", "time": 1, "type": "hour", "price": 14}, {"id": "77642501-5f1d-4757-b277-9df7571637cb", "name": "三小时", "time": 3, "type": "hour", "price": 30}]