Question

我得到的位置结果如下：

10-12-1

并希望得到如下结果：

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？

Answer 1

您可以使用简单的技巧将SUBSTRING_INDEX和GROUP_CONCAT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

的 SqlFiddleDemo

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 ║ ╚═════╩═════════════════╩════════════════╝

请注意，此解决方案适用于长度可变（1-20-300-4000-500000-600000-7）的3个以上的部分。

Answer 2

您可以尝试这样：

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));

<强> FIDDLE DEMO

如何在Mysql中用虚线反转字符串

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