mysql - MySql order by 与今天的日期相关

Question

我有一个日期列，我正在使用 order by 子句。

我想在顶部显示今天日期附近的记录，然后在底部显示过去的所有记录。

我想用单个查询来做到这一点。

我试过这个

SELECT *, DATEDIFF(`date`, CURDATE()) AS diff  FROM `post` order by diff

这个查询的问题是这将首先显示过去的记录，如果我使用降序，那么远离今天的记录将位于顶部。

我怎样才能做到这一点？

任何帮助将不胜感激。

Answer 1

你应该能够做到这一点：

SELECT *, DATEDIFF(`date`, CURDATE()) AS diff  FROM `post`
order by CASE WHEN diff < 0 THEN 1 ELSE 0 END, diff

这将强制过去的任何日期在当前日期或未来日期之后排序。

Answer 2

尝试这个

   SELECT *, DATEDIFF(`date`, CURDATE()) AS diff  FROM `post` order by ABS(diff)

Answer 3

你需要ORDER BY像这样下降：

 SELECT *, DATEDIFF(`date`, CURDATE()) AS diff  
 FROM `post ORDER BY diff DESC;

Answer 4

为什么没有简单的旧日期顺序？

 SELECT *, DATEDIFF(`date`, CURDATE()) AS diff  
 FROM `post ORDER BY date DESC;

Answer 5

由于您不希望将来的结果太远，因此只需从结果集中删除这些结果：

SELECT *, DATEDIFF(`date`, CURDATE()) AS diff FROM `post` 
WHERE `date` < date_add(CURDATE(), INTERVAL 1 DAY)
ORDER BY diff DESC