Question

我遇到以下问题，我想知道如何最好地在SQL中解决它。我有一个如下表：

我想在另一个表中创建如下所示的跨度：

2019-06-24 19:15:05 -> 2019-06-24 23:59:00
2019-06-25 00:00:00 -> 2019-06-26 18:47:39
2019-06-26 19:02:40 -> 2019-06-26 23:59:00 
2019-06-27 00:00:00 -> 2019-06-27 23:59:00 
2019-06-28 00:00:00 -> 2019-06-28 23:59:00 
2019-06-29 00:00:00 -> 2019-06-29 23:59:00 
2019-06-30 00:00:00 -> 2019-06-30 23:59:00
2019-07-01 00:00:00 -> 2019-07-01 23:59:00 
2019-07-02 00:00:00 -> 2019-07-02 01:58:32
2019-07-02 02:01:10 -> 2019-07-02 23:59:00 
2019-07-03 00:00:00 -> 2019-07-03 23:59:00
2019-07-04 00:00:00 -> 2019-07-04 23:59:00
2019-07-05 00:00:00 -> 2019-07-05 01:35:00

因此，基本上，我的跨度需要从FromTimes到ToTimes，并在每天之间的每个午夜使用新行

有人可以让我知道最好的方法吗？我尝试过游标在范围内的每一天进行迭代，但是我认为这不是最好的方法。有人可以给些指导吗？

Answer 1

我不太明白为什么你会在午夜前一分钟停下来。我将使用递归CTE获得完整的一天：

#pragma comment(lib,"pthreadVC2.lib")
#include<stdio.h>
#include<conio.h>
#include<stdlib.h>
#include<pthread.h>
#include<string.h>
#include<time.h>
pthread_t runTheGame;
//***unneccessary code is hidden***
void timerTick()//function for doing countdown
{
    for (;;)
    {
        restTime--;
        if (restTime <= 0)
        {
            pthread_kill(runTheGame,SIGQUIT);//VS doesn't recognize the SIGQUIT
            if (score >= goalOfLevel[arcadeLevel - 1])
            {
                //code is not written yet
            }
        }
        _sleep(1000);
    }
}
//***unneccessary code is hidden***
void gameRunning()//Real game loop. Run by pthread_create().
{
    //***unneccessary code is hidden***
}
//***unneccessary code is hidden***
void game(int arcade)//function for initializing the game
{
    arcadeLevel = arcade;
    boardRange = 4;
    oversize = 2048;
    score = 0;
    revive = 0;
    doubleScoreOrNot = 1;
    if (arcadeLevel > 0)
    {
        restTime = timeOfLevel[arcadeLevel - 1];
        if (passivePower[0] == 1)
        {
            boardRange++;
        }
        if (passivePower[1] == 1)
        {
            revive = 1;
        }
        if (passivePower[3] == 1)
        {
            oversize = 1024;
        }
        if (passivePower[4] == 1)
        {
            score=goalOfLevel[arcadeLevel-1]/10;
        }
        if (passivePower[5] == 1)
        {
            restTime += restTime /20*3;
        }
        if (passivePower[7] == 1)
        {
            boardRange--;
            doubleScoreOrNot = 2;
        }
    }
    pthread_create(&runTheGame, NULL, gameRunning, NULL);//after all these initialization, real game starts here
}

Here是db <>小提琴。

如果可以超过100天，则需要向查询中添加with cte as ( select fromtime, (case when datediff(day, fromtime, totime) = 0 then totime else dateadd(day, 1, convert(date, fromtime)) end) as dayendtime, totime from t union all select dayendtime as fromtime, (case when datediff(day, dayendtime, totime) = 0 then totime else dateadd(day, 1, dayendtime) end) as dayendtime, totime from cte where dayendtime < totime ) select fromtime, dateadd(minute, -1, dayendtime) from cte;。

为一个时间范围内的多行创建午夜到午夜日期范围的行

1 个答案: