Question

我们在MySQL中有一个表，其中一列是date_added。

对于每个date_added，可能有几条记录。

我需要显示一个网页，其中显示按日期降序排序的行，但每个日期最多只有10个。

示例输出：

Jun 1
 item
 item
 item
 item
 item
 item
May 31
 item 
 item etc

有没有办法在MySQL中运行一个可以获取这样结果的查询？

Answer 1

SELECT * FROM records WHERE date_added in (
    SELECT date_added FROM dates ORDER BY date_added DESC
    )

就我的知识有限而言，每个日期不能只选择10行。您需要在应用的业务逻辑中强制执行该限制。

Answer 2

我不确定我到底知道你在追求什么，但你可以尝试：

SELECT * FROM myTable GROUP BY date_added DESC LIMIT 0,10;

Answer 3

据我所知，无法通过查询完成。唯一的解决方案是将此逻辑放入网页的代码中。

确保记录集按日期排序。

在PHP中，一个选项是：

foreach ($rs as $row) 
{
    if ($tmp_date != $row->date_added) { $counter = 0; }

    if ($counter <= 10) {
         echo $row->yourinfo; //Whatever you want echoed from your recordset             
    }

    $counter++;
    $tmp_date = $row->date_added;
}

Answer 4

不确定性能（可能不好）：

SELECT m.* 
FROM 
    myTable AS m
WHERE m.id IN
        ( SELECT ms.id
          FROM myTable AS ms
          WHERE ms.date_added = m.date_added
          ORDER BY ms.something ASC           --- whatever (ASC or DESC)
          LIMIT 10
        )
ORDER BY m.date_added
       , m.something

你也可以试试这个：

SELECT m1.*
FROM myTable AS m1
  JOIN myTable AS m2
    ON m2.date_added = m1.date_added
    AND m2.id <= m1.id                 --- this could also be: 
                                       ---    m2.someField <= m1.someField
GROUP BY m1.id
HAVING COUNT(*) <= 10
ORDER BY m.date_added

或变体：

SELECT m1.*
FROM myTable AS m1
  LEFT JOIN myTable AS m2
    ON m2.date_added = m1.date_added
    AND m2.id < m1.id              
GROUP BY m1.id
HAVING COUNT(*) < 10
ORDER BY m.date_added

Answer 5

SET @rank:=0;
select TTT.f1, TTT.value
from 
(select TT.f1, TT.value,TT.c, 
@rank:=@rank+1 as rnk,  
case when @rank>=TT.c then @rank:=0 end as dummy
from (
select T.f1, T.value, cnt.c
from t2 T, (select f1, count(*) as c from t2 group by f1) cnt
where cnt.f1 = T.f1
order by T.f1 asc, T.value desc) TT ) TTT
where TTT.rnk<=10

其中f1是'date'字段，value是'item'字段

MySQL - 如何从此表中选择并按Date_Added排序？

5 个答案: