有没有一种方法可以将包含json数组的文本列中的数据提取到表中,json数组具有不同数量的json对象?
例如,如果我…
CREATE TABLE tableWithJsonStr (location TEXT, jsonStr TEXT);
INSERT INTO tableWithJsonStr VALUES
('Home', '[{"animalId":"1","type":"dog", "color":"white","isPet":"1"},{"animalId":"2","type":"cat", "color":"brown","isPet":"1"}]'),
('Farm', '[{"animalId":"8","type":"cow", "color":"brown","isPet":"0"}, {"animalId":"33","type":"pig", "color":"pink","isPet":"0"}, {"animalId":"22","type":"horse", "color":"black","isPet":"1"}]'),
('Zoo', '[{"animalId":"5","type":"tiger", "color":"stripes","isPet":"0"}]');
和
CREATE TABLE animal (
location TEXT,
idx INT,
animalId INT,
type TEXT,
color TEXT,
isPet BOOLEAN
);
我可以通过运行提取带有jsonStr.jsonStr的表格
INSERT INTO animal
SELECT location,
idx AS id,
TRIM(BOTH'"' FROM JSON_EXTRACT(jsonStr, CONCAT('$[', idx, '].animalId'))) AS animalId,
TRIM(BOTH'"' FROM JSON_EXTRACT(jsonStr, CONCAT('$[', idx, '].type'))) AS type,
TRIM(BOTH'"' FROM JSON_EXTRACT(jsonStr, CONCAT('$[', idx, '].color'))) AS color,
TRIM(BOTH'"' FROM JSON_EXTRACT(jsonStr, CONCAT('$[', idx, '].isPet'))) AS isPet
FROM tableWithJsonStr
JOIN(
SELECT 0 AS idx UNION
SELECT 1 AS idx UNION
SELECT 2 AS idx UNION
SELECT 3 AS idx
) AS indexes
WHERE JSON_EXTRACT(jsonStr, CONCAT('$[', idx, ']')) IS NOT NULL;
动物表的结果是:
| location | idx | animalId | type | color | isPet |
|==========|=====|==========|=======|=========|=======|
| Farm | 0 | 8 | cow | brown | 0 |
| Farm | 1 | 33 | pig | pink | 0 |
| Farm | 2 | 22 | horse | black | 1 |
| Home | 0 | 1 | dog | white | 1 |
| Home | 1 | 2 | cat | brown | 1 |
| Zoo | 0 | 5 | tiger | stripes | 0 |
虽然该解决方案有效,但它是不可扩展的。如果我的json数组中有3个以上的对象,除非我在JOIN中添加另一个SELECT 4 AS idx,否则它们将不会被计算在内。有没有更好的方法可以迭代数组中的对象,而不需要预先知道每个数组中可能存在的最大对象数?
如果您使用的是MySQL 8.0,您可以使用JSON_TABLE
命令从JSON
:的每一行提取数据
SELECT t1.location, farm.*
FROM tableWithJsonStr t1
JOIN JSON_TABLE(t1.jsonStr,
'$[*]'
COLUMNS (idx FOR ORDINALITY,
animalId INT PATH '$.animalId',
type TEXT PATH '$.type',
color TEXT PATH '$.color',
isPet BOOLEAN PATH '$.isPet')
) farm
ORDER BY location, idx
输出:
location idx animalId type color isPet
Farm 1 8 cow brown 0
Farm 2 33 pig pink 0
Farm 3 22 horse black 1
Home 1 1 dog white 1
Home 2 2 cat brown 1
Zoo 1 5 tiger stripes 0
dbfiddle 演示
如果你被MySQL 5.7卡住了,你可以使用一个存储过程来提取数据:
DELIMITER $$
CREATE PROCEDURE extract_animals()
BEGIN
DECLARE idx INT;
DECLARE finished INT DEFAULT 0;
DECLARE loc, json VARCHAR(200);
DECLARE json_cursor CURSOR FOR SELECT location, jsonStr FROM tableWithJsonStr;
DECLARE CONTINUE HANDLER FOR NOT FOUND SET finished = 1;
DROP TABLE IF EXISTS animal;
CREATE TABLE animal (location TEXT, idx INT, animalId INT, type TEXT, color TEXT, isPet BOOLEAN);
OPEN json_cursor;
json_loop: LOOP
FETCH json_cursor INTO loc, json;
IF finished = 1 THEN
LEAVE json_loop;
END IF;
SET idx = 0;
WHILE JSON_CONTAINS_PATH(json, 'one', CONCAT('$[', idx, ']')) DO
INSERT INTO animal VALUES(loc,
idx,
JSON_UNQUOTE(JSON_EXTRACT(json, CONCAT('$[', idx, '].animalId'))),
JSON_UNQUOTE(JSON_EXTRACT(json, CONCAT('$[', idx, '].type'))),
JSON_UNQUOTE(JSON_EXTRACT(json, CONCAT('$[', idx, '].color'))),
JSON_UNQUOTE(JSON_EXTRACT(json, CONCAT('$[', idx, '].isPet'))));
SET idx = idx + 1;
END WHILE;
END LOOP json_loop;
END $$
输出:
location idx animalId type color isPet
Home 0 1 dog white 1
Home 1 2 cat brown 1
Farm 0 8 cow brown 0
Farm 1 33 pig pink 0
Farm 2 22 horse black 1
Zoo 0 5 tiger stripes 0
dbfiddle 演示