select * from test as 'time' where time.update_time<DATE_SUB(CURDATE(), INTERVAL 2 MONTH) 查询上一年的数据 select * from test as 'time' where time.update_time<DATE_SUB(CURDATE(), INTERVAL 1 YEAR) 查询距离当前现在6个月的数据 select name,submittime from enterprise where submittime between...
selectname,submittimefromenterprisewheredate_format(submittime,'%Y-%m')=date_format(DATE_SUB(curdate(),INTERVAL1MONTH),'%Y-%m')select*fromuserwhereDATE_FORMAT(pudate,'%Y%m')=DATE_FORMAT(CURDATE(),'%Y%m') ;select*fromuserwhereWEEKOFYEAR(FROM_UNIXTIME(pudate,'%y-%m-%d'))=WEEKOFYEAR(now()...
可以使用以下代码来实现: # 将日期字符串转换为日期对象fromdatetimeimportdatetime start_date_obj=datetime.strptime(start_date,"%Y-%m-%d")end_date_obj=datetime.strptime(end_date,"%Y-%m-%d")# 计算月份差month_diff=(end_date_obj.year-start_date_obj.year)*12+(end_date_obj.month-start_date_obj...
select name,submittime from enterprise where date_format(submittime,’%Y-%m’)=date_format(DATE_SUB(curdate(), INTERVAL 1 MONTH),’%Y-%m’) select*from`user`whereDATE_FORMAT(pudate,‘%Y%m‘)=DATE_FORMAT(CURDATE(),‘%Y%m‘) ; select * from user where WEEKOFYEAR(FROM_UNIXTIME(pudate,’...
快速给月份、日期前面补0 String(new Date().getMonth()+1).padStart(2,0) 获得月份，此时类型为...
YEAR(date)返回日期date的年份(1000~9999) 一些示例: 获取当前系统时间: 代码语言:javascript 复制 SELECTFROM_UNIXTIME(UNIX_TIMESTAMP());SELECTEXTRACT(YEAR_MONTHFROMCURRENT_DATE);SELECTEXTRACT(DAY_SECONDFROMCURRENT_DATE);SELECTEXTRACT(HOUR_MINUTEFROMCURRENT_DATE); ...
MONTH——获取指定日期中的月份。 WEEK——获取指定日期是一年中的第几周。 YEAR——获取年份。 QUARTER——获取日期所在的季度值。 DATE_ADD 和 ADDDATE——两个函数功能相同,都是向日期添加指定的时间间隔。 DATE_SUB 和 SUBDATE——两个函数功能相同,都是向日期减去指定的时间间隔。
EXTRACT(type FROM date)函数所使用的时间间隔类型说明符同DATE_ADD()或DATE_SUB()的相同,但它是从日期中提取一部分,而不是执行日期运算。案例:使用EXTRACT()函数提取日期值或时间值,SQL语句如下: 命令语句:selectEXTRACT(YEAR FROM ‘2018-08-12’) as col1, EXTRACT(YEAR_MONTH FROM ‘2018-08-12’) as...
Date: October 24, 2012 10:57AM The idea is to turn year-month comparison into simple arithmetic ... select count(*) from ordertable where 100*year(order_date)+month(order_date) = 100*year(curdate())+month(curdate()); Edited 1 time(s). Last edit at 10/24/2012 11:29AM by Peter...
Date: October 24, 2012 03:32PM So this: select count(*) from ordertable where 100*year(order_date)+month(order_date) = 100*year(curdate())+month(curdate()); would find: the number of orders from the previous month? OR the number of customers who had at least one order in the pr...