我在MySQL中有一个表,它有一个存储JSON对象的列。 如何轻松运行WHERE子句中可能包含某些JSON字段的查询?
EX:
使用名为articles
+----+---------+--------------------------------------------------------------------------------------------------+
| id | user_id | json_data |
+----+---------+--------------------------------------------------------------------------------------------------+
| 1 | 1 | {"url":"https://www.cpubenchmark.net/","title": "CPU Benchmarks"} |
| 2 | 1 | {"url":"http://www.ebay.com/sch/CPUs-Processors-/164/i.html","title": "Computer and Processors"} |
| 3 | 2 | {"url":"https://www.youtube.com/watch?v=tntOCGkgt98","title": "Funny Cats Compilation" |
+----+---------+--------------------------------------------------------------------------------------------------+
我希望能够写出如下内容:
SELECT user_id, json_data FROM articles WHERE json_data.title LIKE "%CPU%"
那应该只返回第一行。
答案 0 :(得分:8)
您可以使用json_extract
。
https://dev.mysql.com/doc/refman/5.7/en/json-search-functions.html#function_json-extract
SELECT user_id, json_data
FROM articles
WHERE json_extract(json_data, '$.title') LIKE '%CPU%';
答案 1 :(得分:5)
我最终以这种方式解决:https://github.com/ChrisCinelli/mysql_json使用UDF。我详细介绍了如何编译它并在README中安装它。它适用于我Ubuntu 12.04.5
gcc version 4.6.3
MySQL 5.5
您将能够运行:
SELECT json_get('{"a":1}', 'a') => 1
SELECT json_get('{"a":1}', 'b') => NULL
SELECT json_get('[1,2,3]', 2) => 3
SELECT json_get('{"a":[2]}', 'a', 0) => 2
#Also:
SELECT json_get('{"a":{"b":2}}', 'a') => object
SELECT json_get('{"a":[1,2,3]}', 'a') => array
# Verify if it is a valid JSON:
SELECT ISNULL(json_get('{"a":1}')); => 0 # Valid
SELECT ISNULL(json_get('{"a":1')); => 1 # Invalid
# Create an example table:
CREATE TABLE `message` (
`id` int(11) NOT NULL AUTO_INCREMENT,
`data` text,
PRIMARY KEY (`id`)
);
INSERT INTO message (id,data) VALUES(1,'{"from":"chris","title":"Awesome Article","body":"Lorem ipsum dolor sit amet, consectetur adipiscing elit."}');
INSERT INTO message (id,data) VALUES(2,'{"from":"loren","title":"Another Article","body":"Lorem ipsum dolor sit amet, consectetur adipiscing elit."}');
INSERT INTO message (id,data) VALUES(3,'{"from":"jason","title":"How to run a query","body":"Lorem ipsum dolor sit amet, consectetur adipiscing elit."}');
# Run queries on JSON values:
SELECT json_get(data,'title') FROM message WHERE id=2;
SELECT id,data FROM message WHERE json_get(data,'from')='chris';
SELECT id,data FROM message WHERE json_get(data,'title') LIKE '%Article%';
答案 2 :(得分:2)
尝试以下查询,看看它是否符合您的需求:
SELECT user_id, json_data
FROM articles
WHERE common_schema.extract_json_value(json_data,'title')
LIKE "%CPU%"
这仅适用于MySQL
版本5.1或更新版本。