这不是工作,DB
但有可能:
CREATE TABLE tab(id INT, col VARCHAR(100));
INSERT INTO tab(id, col)
VALUES (1, 'option[A]sum[A]g3et[B]'), (2, '[Cosi]sum[A]g3et[ZZZZ]');
SELECT DISTINCT *
FROM (
SELECT id, RIGHT(val, LENGTH(val) - LOCATE('[', val)) AS val
FROM
(
SELECT id, SUBSTRING_INDEX(SUBSTRING_INDEX(t.col, ']', n.n), ']', -1) AS val
FROM tab t
CROSS JOIN
(
SELECT a.N + b.N * 10 + 1 n
FROM
(SELECT 0 AS N UNION ALL SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3 UNION ALL SELECT 4 UNION ALL SELECT 5 UNION ALL SELECT 6 UNION ALL SELECT 7 UNION ALL SELECT 8 UNION ALL SELECT 9) a
,(SELECT 0 AS N UNION ALL SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3 UNION ALL SELECT 4 UNION ALL SELECT 5 UNION ALL SELECT 6 UNION ALL SELECT 7 UNION ALL SELECT 8 UNION ALL SELECT 9) b
) n
WHERE n.n <= 1 + (LENGTH(t.col) - LENGTH(REPLACE(t.col, ']', '')))
) sub
) s
WHERE val <> ''
ORDER BY ID;
笔记:
根据col最大长度,您可能需要在CROSS JOIN部分中生成更多数字。现在最多可以有100个。
输出:
在此处输入图片说明
这个怎么运作:
最内层查询:
╔════╦══════════╗
║ id ║ val ║
╠════╬══════════╣
║ 1 ║ option[A ║
║ 1 ║ sum[A ║
║ 1 ║ g3et[B ║
║ 1 ║ ║
╚════╩══════════╝
第二个子查询:
╔════╦═════╗
║ id ║ val ║
╠════╬═════╣
║ 1 ║ A ║
║ 1 ║ A ║
║ 1 ║ B ║
║ 1 ║ ║
╚════╩═════╝
最外层的查询:
╔════╦═════╗
║ id ║ val ║
╠════╬═════╣
║ 1 ║ A ║
║ 1 ║ B ║
╚════╩═════╝
我需要每行查询的结果..不合并
因此添加简单:
WHERE n.n <= 1 + (LENGTH(t.col) - LENGTH(REPLACE(t.col, ']', '')))
AND t.id = ?
编辑2:
请参阅http://sqlfiddle.com/#!9/8ee95/1,您的查询对我的数据有部分作用。我也将类型更改为longtext。
您想在MysqL中解析JSON。正如我在解析并在应用程序层中获取价值之前所说的那样。该答案仅用于演示/玩具目的,并且性能会很差。
SELECT id, val,s.n
FROM (
SELECT id, RIGHT(val, LENGTH(val) - LOCATE('[', val)) AS val,n
FROM
(
SELECT id, SUBSTRING_INDEX(SUBSTRING_INDEX(t.col, ']', n.n), ']', -1) AS val, n.n
FROM (SELECT id, REPLACE(col, '[]','') as col FROM tab) t
CROSS JOIN
(
SELECT e.N * 10000 + d.N * 1000 + c.N * 100 + a.N + b.N * 10 + 1 n
FROM
(SELECT 0 AS N UNION ALL SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3 UNION ALL SELECT 4 UNION ALL SELECT 5 UNION ALL SELECT 6 UNION ALL SELECT 7 UNION ALL SELECT 8 UNION ALL SELECT 9) a
,(SELECT 0 AS N UNION ALL SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3 UNION ALL SELECT 4 UNION ALL SELECT 5 UNION ALL SELECT 6 UNION ALL SELECT 7 UNION ALL SELECT 8 UNION ALL SELECT 9) b
,(SELECT 0 AS N UNION ALL SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3 UNION ALL SELECT 4 UNION ALL SELECT 5 UNION ALL SELECT 6 UNION ALL SELECT 7 UNION ALL SELECT 8 UNION ALL SELECT 9) c
,(SELECT 0 AS N UNION ALL SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3 UNION ALL SELECT 4 UNION ALL SELECT 5 UNION ALL SELECT 6 UNION ALL SELECT 7 UNION ALL SELECT 8 UNION ALL SELECT 9) d
,(SELECT 0 AS N UNION ALL SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3 UNION ALL SELECT 4 UNION ALL SELECT 5 UNION ALL SELECT 6 UNION ALL SELECT 7 UNION ALL SELECT 8 UNION ALL SELECT 9) e
) n
WHERE n.n <= 1 + (LENGTH(t.col) - LENGTH(REPLACE(t.col, ']', '')))
) sub
) s
WHERE val <> ''
GROUP BY id, val
HAVING n <> MAX(n)
ORDER BY id,n;
sqlfiddleDemo
输出:
╔═════╦═════════════╦════╗
║ id ║ val ║ n ║
╠═════╬═════════════╬════╣
║ 1 ║ CE31285LV4 ║ 1 ║
║ 1 ║ D32E ║ 3 ║
║ 1 ║ GTX750 ║ 5 ║
║ 1 ║ M256S ║ 7 ║
║ 1 ║ H2X1T ║ 9 ║
║ 1 ║ FMLANE4U4 ║ 11 ║
╚═════╩═════════════╩════╝
编辑3:
在那里到底做了什么?你为什么需要
CROSS JOIN并且整个子查询仅是理货表格。就这些。如果MysqL具有产生数字序列的功能(如generate_series或预先填充的数字表),则不需要CROSS JOIN。
需要以下数字表SUBSTRING_INDEX:
SUBSTRING_INDEX(str,delim,count)
在出现定界符delim之前,从字符串str返回子字符串。如果count为正,则返回最后定界符左侧的所有内容(从左侧开始计数)。如果count为负,则返回最后定界符右边的所有内容(从右边开始计数)。搜索delim时,SUBSTRING_INDEX()执行区分大小写的匹配。