获取 mySQL MONTH() 以使用前导零？

Question

我如何指定 mySQL 的 MONTH() function 在此查询中返回 '08' 而不是 8？

我希望按日期排序。 目前正在获取日期的结果，例如

2006-9
2007-1
2007-10
2007-11

当前查询：

SELECT COUNT(*), CONCAT(YEAR(`datetime_added`), '-', MONTH(`datetime_added`)) as date FROM `person` WHERE (email = '' OR email IS NULL) 
GROUP BY date 
ORDER BY date ASC

Answer 1

请改用以下内容：

DATE_FORMAT(`datetime_added`,'%Y-%m')

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解释：

DATE_FORMAT() function 允许您使用下表中描述的说明符来格式化日期（从文档中逐字记录）。 因此，格式字符串'%Y-%m'意思是：“一整年（4 位数字），后跟一个破折号（ - ），后跟一个两位数的月份编号”。

请注意，您可以通过设置lc_time_names系统变量来指定用于日/月名称的语言。 非常有用。 有关更多详细信息，请参阅文档。

Specifier   Description
%a  Abbreviated weekday name (Sun..Sat)
%b  Abbreviated month name (Jan..Dec)
%c  Month, numeric (0..12)
%D  Day of the month with English suffix (0th, 1st, 2nd, 3rd, …)
%d  Day of the month, numeric (00..31)
%e  Day of the month, numeric (0..31)
%f  Microseconds (000000..999999)
%H  Hour (00..23)
%h  Hour (01..12)
%I  Hour (01..12)
%i  Minutes, numeric (00..59)
%j  Day of year (001..366)
%k  Hour (0..23)
%l  Hour (1..12)
%M  Month name (January..December)
%m  Month, numeric (00..12)
%p  AM or PM
%r  Time, 12-hour (hh:mm:ss followed by AM or PM)
%S  Seconds (00..59)
%s  Seconds (00..59)
%T  Time, 24-hour (hh:mm:ss)
%U  Week (00..53), where Sunday is the first day of the week
%u  Week (00..53), where Monday is the first day of the week
%V  Week (01..53), where Sunday is the first day of the week; used with %X
%v  Week (01..53), where Monday is the first day of the week; used with %x
%W  Weekday name (Sunday..Saturday)
%w  Day of the week (0=Sunday..6=Saturday)
%X  Year for the week where Sunday is the first day of the week, numeric, four digits; used with %V
%x  Year for the week, where Monday is the first day of the week, numeric, four digits; used with %v
%Y  Year, numeric, four digits
%y  Year, numeric (two digits)
%%  A literal “%” character
%x  x, for any “x” not listed above 

Answer 2

您可以使用填充

SELECT
    COUNT(*), 
    CONCAT(YEAR(`datetime_added`), '-', LPAD(MONTH(`datetime_added`), 2, '0')) as date 
FROM `person` 
WHERE (email = '' OR email IS NULL) 
GROUP BY date 
ORDER BY date ASC

Answer 3

DATE_FORMAT(`datetime_added`,'%Y - %m')

Answer 4

MONTH() 返回一个 integer，所以当然没有前导零。 您需要将其转换为字符串，左填充“0”并取最后 2 个字符。

Answer 5

使用LPAD function 也可以

LPAD(MONTH(your_date_column_here),2,'0')

请参阅下面使用当前日期的示例

Answer 6

一个月

SELECT DATE_FORMAT('2023-2-1','%m'); -- 02

一天

SELECT DATE_FORMAT('2023-2-1','%d');  -- 01