查找MySQL JSON对象或数组的交集 [英] Find intersection of MySQL JSON objects or arrays

问题描述

问题与 MySQL/MariaDB JSON函数有关.
如何找到多个JSON结构的交集?
在PHP中，可以使用以下代码完成:

The question is about MySQL/MariaDB JSON Functions.
How do you find intersection of multiple JSON structures?
In PHP it is done using this code:

array_intersect(
    ['a', 'b'],
    ['b', 'c']
);

如果我们想象一个名为JSON_INTERSECT的函数，代码将如下所示:

If we imagine a function named JSON_INTERSECT, the code would look like this:

SET @json1 = '{"key1": "a", "key2": "b"}';
SET @json2 = '["b", "c"]';
SET @json3 = '["c", "d"]';

SELECT JSON_INTERSECT(@json1, @json2); // returns '["b"]';
SELECT JSON_INTERSECT(@json1, @json3); // returns NULL;

推荐答案

看来，没有很好的内置方法可以做到这一点，并且在该主题上仍然没有好的答案，所以我想我应该快速补充一下&肮脏的解决方案.如果执行以下代码，它将创建一个名为MY_JSON_INTERSECT的函数，该函数将输出与指定的原始海报完全相同的结果.在信任我的代码之前，请确保您已仔细检查并可以创建一个新函数:

It looks there are no good built-in ways of doing this and there are still no good answers on this topic, so I thought I'd add my quick & dirty solution. If you execute the below code it will create a function called MY_JSON_INTERSECT that will output results exactly as the original poster specified. Make sure you've looked this over and are ok with creating a new function before trusting my code:

delimiter $$
CREATE FUNCTION `MY_JSON_INTERSECT`(Array1 VARCHAR(1024), Array2 VARCHAR(1024)) RETURNS varchar(1024)
BEGIN
    DECLARE x int;
    DECLARE val, output varchar(1024);
    SET output = '[]';
    SET x = 0;
    IF JSON_LENGTH(Array2) < JSON_LENGTH(Array1) THEN
        SET val = Array2;
        SET Array2 = Array1;
        SET Array1 = val;
    END IF;
    WHILE x < JSON_LENGTH(Array1) DO
        SET val = JSON_EXTRACT(Array1, CONCAT('$[',x,']'));
        IF JSON_CONTAINS(Array2,val) THEN
            SET output = JSON_MERGE(output,val);
        END IF;
        SET x = x + 1; 
    END WHILE;
    IF JSON_LENGTH(output) = 0 THEN
        RETURN NULL;
    ELSE
        RETURN output;
    END IF;
END$$

然后您可以像这样调用函数:

You can then call the function like this:

SELECT MY_JSON_INTERSECT('[1,2,3,4,5,6,7,8]','[0,3,5,7,9]');

输出:

[3,5,7]

这不是美观或有效的方法，但它确实有效……希望很快能有更好的答案.

This isn't beautiful or efficient, but it's something that works... Hopefully better answers will come soon.

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