MySQL ORDER BY 字符串作为数字（较大的数字在较小的数字之上）

Question

考虑以下结果，其中 code 是类型 VARCHAR:

SELECT code
FROM lock
ORDER BY CAST(code AS signed) > 0 DESC, `code

|code|
|4   |  
|420 |
|5   |
|T6  |
|X30 |

如何更改查询，使其 returns 结果按以下顺序排列：

|code|
|4   |  
|5   |
|420 |
|T6  |
|X30 |

Answer 1

w=12=WILL.y.:w=11=w w=10=sh

Answer 2

SELECT code
FROM lock
ORDER BY CAST(code AS signed) > 0 DESC, CAST(code AS signed) ASC, code ASC

第一个顺序会将数字排到前面。第二个顺序只会对数字进行升序排序，而第一个顺序会将它们保留在开头。第三个将仅对字符串进行升序排序，数字将保持其顺序，因为它们已经排序。

Answer 3

LENGTH()SQL中的函数用于获取字符串的长度：

SELECT code
FROM lock
ORDER BY LENGTH(code), code

输出：

+------+
| code |
+------+
| 4    |
| 5    |
| T6   |
| 420  |
| X30  |
+------+

MySQL ORDER BY string as number (larger numbers above smaller numbers)