如何通过 Mysql 中的破折号反转字符串
How to reverse string by dashed in Mysql
我得到的排名结果如下:
10-12-1
并希望得到这样的结果:
1-12-10
我正在应用字符串函数 REVERSE()
而不是结果
query : select REVERSE("10-12-1")
result : 1-21-01
我该如何解决?
您可以使用简单的技巧将其与 SUBSTRING_INDEX
和 GROUP_CONCAT
一起解析回一个字符串:
SELECT id, col, GROUP_CONCAT(val ORDER BY n DESC SEPARATOR '-') AS reversed
FROM
(
SELECT id, col, SUBSTRING_INDEX(SUBSTRING_INDEX(t.col, '-', n.n), '-', -1) AS val, n
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
GROUP BY id
输出:
╔═════╦═════════════════╦════════════════╗
║ id ║ col ║ reversed ║
╠═════╬═════════════════╬════════════════╣
║ 1 ║ 1-12-10 ║ 10-12-1 ║
║ 2 ║ 111-12-10 ║ 10-12-111 ║
║ 3 ║ 11123-2-103223 ║ 103223-2-11123 ║
╚═════╩═════════════════╩════════════════╝
请记住,此解决方案适用于 3 个以上的可变长度部分 (1-20-300-4000-500000-600000-7
)。
你可以这样试试:
select concat(SUBSTRING_INDEX('10-12-1', '-', -1) , '-'
, substr('10-12-1',instr('10-12-1',"-") + 1, instr('10-12-1',"-"))
, LEFT('10-12-1',LOCATE('-','10-12-1') - 1));
我得到的排名结果如下:
10-12-1
并希望得到这样的结果:
1-12-10
我正在应用字符串函数 REVERSE()
而不是结果
query : select REVERSE("10-12-1")
result : 1-21-01
我该如何解决?
您可以使用简单的技巧将其与 SUBSTRING_INDEX
和 GROUP_CONCAT
一起解析回一个字符串:
SELECT id, col, GROUP_CONCAT(val ORDER BY n DESC SEPARATOR '-') AS reversed
FROM
(
SELECT id, col, SUBSTRING_INDEX(SUBSTRING_INDEX(t.col, '-', n.n), '-', -1) AS val, n
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
GROUP BY id
输出:
╔═════╦═════════════════╦════════════════╗
║ id ║ col ║ reversed ║
╠═════╬═════════════════╬════════════════╣
║ 1 ║ 1-12-10 ║ 10-12-1 ║
║ 2 ║ 111-12-10 ║ 10-12-111 ║
║ 3 ║ 11123-2-103223 ║ 103223-2-11123 ║
╚═════╩═════════════════╩════════════════╝
请记住,此解决方案适用于 3 个以上的可变长度部分 (1-20-300-4000-500000-600000-7
)。
你可以这样试试:
select concat(SUBSTRING_INDEX('10-12-1', '-', -1) , '-'
, substr('10-12-1',instr('10-12-1',"-") + 1, instr('10-12-1',"-"))
, LEFT('10-12-1',LOCATE('-','10-12-1') - 1));