用SQL语句计算出给定日期是星期几？

Posted 2023-04-27

参考技术A

1、创建测试表，

create table test_day(id number, v_date date);

2、插入测试数据

insert into test_day 

select level, sysdate-level from dual connect by level < 8; 

3、查询表中记录，select t.*, rowid from test_day t;

4、编写sql，查看表中每条记录日期对应的周几，

c语言编程题：输入一个日期，计算出该日期是星期几？

c语言编程题：输入一个日期，计算出该日期是星期几？

源代码如下：

/*
* author:X
* date:2021/9/17 18:49
* description:输入一个日期，计算出该日期是星期几？
*/

#include <stdio.h>
int leap(int year);
int main()
{
    int year, month, day, i;
    int m[12] = { 31,28,31,30,31,30,31,31,30,31,30,31 };
    printf("请输入要判断的年月日(格式如:2021 9 17 年月日之间用空格隔开)：");
    while (~scanf("%d %d %d", &year, &month, &day))
    {
        if (day < 1 || month < 1 || month > 12 || year < 1 || year>10000|| day > m[month - 1])
        {
            printf("%d/%d/%d 是不正确的日期\\n", year, month, day);
            return -1;
        }
        if (leap(year))
            m[1] = 29;
        else
            m[1] = 28;
        int sum = 0;
        sum += day;
        for (i = 1; i < month; i++)
            sum += m[i - 1];
        for (i = 1; i < year; i++)
        {
            sum += 365;
            if (leap(i))
                sum++;
        }
        int  dx = sum % 7;
        if (dx == 0)
            printf("%d/%d/%d 是星期天\\n", year, month, day);
        else if (dx == 1)
            printf("%d/%d/%d 是星期一\\n", year, month, day);
        else if (dx == 2)
            printf("%d/%d/%d 是星期二\\n", year, month, day);
        else if (dx == 3)
            printf("%d/%d/%d 是星期三\\n", year, month, day);
        else if (dx == 4)
            printf("%d/%d/%d 是星期四\\n", year, month, day);
        else if (dx == 5)
            printf("%d/%d/%d 是星期五\\n", year, month, day);
        else if (dx == 6)
            printf("%d/%d/%d 是星期六\\n", year, month, day);
    }
    return 0;
}

int leap(int year)                  //判断是否为闰年
{
    if (year % 400 == 0 || (year % 4 == 0 && year % 100 != 0))
        return 1;
    return 0;
}

测试结果：