Question

    date      action 
----------   -------
2018-01-12   acquire
2018-01-14   release
2018-01-15   acquire
2018-01-19   release

我需要SELECT WHERE date＆gt; =＆＃39; 2018-01-13＆＃39;但只有一些人才能获得＆＃39;行（按日期ASC排序），结果将是

2018-01-15   acquire
2018-01-19   release

（从满足日期条件后的第一次获得）。是否可以使用单个SQL语句？

Answer 1

获取满足条件的最小日期，然后选择日期开始的所有行。

SELECT date, action
FROM yourTable
WHERE date >= (
    SELECT MIN(date)
    FROM yourTable
    WHERE date >= '2018-01-13' AND action = 'acquire'
)

Answer 2

//Just modifying the texture name to the right format
wstring tempName(fileName);
m_name = string(tempName.begin(), tempName.end());
//Modifying texture name (From original loaded textures names)
int pos = m_name.find_last_of("/");
if(pos >= 0)
{
    m_name = m_name.substr(pos + 1, m_name.length());
}
m_name = m_name.substr(0, m_name.find_last_of("."));

//load the texture
result = CreateWICTextureFromFile(device, fileName, NULL, &m_texture, NULL);

Answer 3

假设您正在尝试获取每个操作的第一个日期......

SELECT `date`, `action`
FROM theTable
WHERE (`date`, `action`) IN (
   SELECT MIN(`date`), `action`
   FROM theTable
   WHERE `date` > '2018-01-13'
   GROUP BY `action`
)

根据订购结果选择条件+另一个条件

3 个答案: