表TESTER 字段:id -- INT date -- TIMESTAMP 1.如何按年.月.日分组查询? select DATE_FORMAT(date,'%Y-%m-%d') time, count(*) count from TESTER group by year(date), month(date), day(date); 其中year().month().day()分别是提取date中的年.月.日字段. 2.时间分组查询的效率? 在不建立索引时,我100W行数据进行测试,用
package com.hanqi.test; public class Clock { int hh; int mm; int ss; String time; Clock(int h,int m,int s) { hh=h; mm=m; ss=s; time=hh+"时"+mm+"分"+ss+"秒"; } String show() { return time; } } package com.hanqi.test; public class
实验现象: 核心代码: int main(void) { /* USER CODE BEGIN 1 */ RTC_TimeTypeDef sTime; RTC_DateTypeDef sDate; ; /* USER CODE END 1 */ /* MCU Configuration----------------------------------------------------------*/ /* Reset of all peripherals, Initializes the F
实验现象: 核心代码: int main(void) { /* USER CODE BEGIN 1 */ RTC_DateTypeDef sDate; RTC_TimeTypeDef sTime; uint8_t second_tmp = ; /* USER CODE END 1 */ /* MCU Configuration----------------------------------------------------------*/ /* Reset of all periphera
1.按年 select to_char(record_date,'yyyy'), sum(col_8) as total_money from table_name where group by to_char(record_date,'yyyy') 2.按月 select to_char(record_date,'yyyy-mm'), sum(col_8) as total_money from table_name where group by to_char(record_date,'yy
select DATE_FORMAT(f.upload_time,'%Y%u') weeks,count(*),sum(p.download_times),sum(p.collection_times),sum(p.click_times) from file_base f left join file_property p on f.property_id = p.id group by weeks; select DATE_FORMAT(f.upload_time,'%Y-%m-%d') d