我正在尝试了解“新的” MYSQL JSON字段。
我有这张桌子:
id (int-11, not_null, auto_inc)
customer_id (int-11, not null)
labels (json)
具有以下数据:
id: 1
customer_id: 1
labels: [{"isnew": "no", "tagname": "FOO", "category": "CAT_1", "isdeleted": "no"}, {"isnew": "yes", "tagname": "BAR", "category": "CAT_2", "isdeleted": "no"}]
JSON美化
[
{
"tagname": "FOO",
"category": "CAT_1",
"isnew": "no",
"isdeleted": "no"
},
{
"tagname": "BAR",
"category": "CAT_2",
"isnew": "yes",
"isdeleted": "no"
}
]
现在我想在表中选择所有具有特定类别和特定标记名
的客户(按customer_id)我尝试了这个:
SELECT * FROM labels_customers_json
WHERE JSON_SEARCH(labels, 'all', 'BAR') IS NOT NULL
但这不是我想要的。这是在每个json属性中搜索。 我已经看到了JSON_EXTRACT的一些示例:
SELECT * FROM `e_store`.`products`
WHERE
`category_id` = 1
AND JSON_EXTRACT(`attributes` , '$.ports.usb') > 0
AND JSON_EXTRACT(`attributes` , '$.ports.hdmi') > 0;
SELECT c, c->"$.id", g, n
FROM jemp
WHERE JSON_EXTRACT(c, "$.id") > 1
ORDER BY c->"$.name";
所以我尝试了
SELECT * FROM labels_customers_json
WHERE JSON_EXTRACT(labels, '$.tagname') = 'BAR'
SELECT labels, JSON_EXTRACT(labels, "$.customer_id"), customer_id
FROM labels_customers_json
WHERE JSON_EXTRACT(labels, "$.customer_id") > 0
答案 0 :(得分:2)
您可能可以尝试使用SELECT * FROM labels_customers_json WHERE JSON_SEARCH(labels, 'all', "BAR", NULL, "$[*].tagname") is not null
-尽管我不能说这是否是执行此查询的最佳方法。
答案 1 :(得分:1)
您也可以使用JSON_SEARCH
搜索特定路径。因此,您可以使用以下查询:
SELECT *
FROM labels_customers_json
WHERE JSON_SEARCH(labels, 'all', 'BAR', NULL, '$[*].tagname') IS NOT NULL
您还可以同时使用JSON_EXTRACT
和JSON_CONTAINS
:
SELECT *
FROM labels_customers_json
WHERE JSON_CONTAINS(JSON_EXTRACT(labels, '$[*].tagname'), '["BAR"]') > 0;
您也只能使用JSON_CONTAINS
进行检查:
SELECT *
FROM labels_customers_json
WHERE JSON_CONTAINS(labels, '{"tagname":"BAR"}') > 0;