select nnd as '年龄段',count(*) as '人数' from( select case when age>= and age<= then '1-10' when age>= and age<= then '11-20' when age>= and age<= then '21-30' when age>= and age<= then '31-40' when age>= and age<= then '41-50' w
Share 一个昨天写的函数. 目的是求给定的时间是所在月份的第几个礼拜. DELIMITER $$ USE `t_girl`$$ DROP FUNCTION IF EXISTS `weekofmonth`$$ CREATE DEFINER=`root`@`localhost` FUNCTION `weekofmonth`( f_datetime DATETIME ) RETURNS INT(11) BEGIN -- Created by ytt. DECLARE v_result INT; D
SELECT id,lng,lat,ROUND(6378.138*2*ASIN(SQRT(POW(SIN((lat1*PI()/180-lat*PI()/180)/2),2)+COS(lat1*PI()/180)*COS(lat*PI()/180)*POW(SIN((lng1*PI()/180-lng*PI()/180)/2),2)))*1000)AS juliFROM address having juli > 500ORDER BY juli AscLIMIT 100 lng1为自定义的经度
mysql数据库获取年龄:TIMESTAMPDIFF(YEAR, [出生日期字段], CURDATE()) select * from (select name 姓名,TIMESTAMPDIFF(YEAR, [出生日期字段], CURDATE()) 年龄 from [表名] )a; sqlserver数据库获取年龄:DATEDIFF(yy,[出生日期字段],GETDATE()) select id AS 编号,SStudentName AS 姓名,DATEDIFF(yy,[出生日期字段],G