如何最小化我的查询？

Question

我希望尽量减少我的查询

"SELECT * FROM tb_videos GROUP BY DATE(added) ORDER BY DATE(added) desc" 

GET结果

foreach($result as $rst) 
{  
    $dt=date('Y-m-d',strtotime($rst->added)); 
    SELECT * FROM tb_videos WHERE DATE(added)='$dt' ORDER BY added desc 
} 

我可以做一个查询后

？

Answer 1

您想最小化找到的记录吗？使用where子句来选择具体的记录

“SELECT * FROM tb_videos WHERE videogroup =动作GROUP BY DATE（加）ORDER BY DATE（加）desc” 的例子

对于

Answer 2

SELECT * FROM tb_videos WHERE DATE(added) = curdate() ORDER BY added desc 

Answer 3

你查询没有意义。首先，您要从数据中选择DISTINCT日期（忽略时间）。那么对于每个日期，你选择该日期的所有数据？

你为什么不刚刚火一个查询

SELECT *, DATE(added) as DateNoTime 
FROM tb_videos 
ORDER BY added desc 

如果你只想要5个日期和表很大，有两种可能性。

1，有没有在日期的差距，您可以使用

SELECT *, DATE(added) as DateNoTime 
FROM tb_videos 
WHERE added >= ADDDATE(CURDATE(), interval -4 days) 
ORDER BY added desc 

2，如果有可能的差距，例如没有为昨天所以它必须显示最近5天有记录

SELECT *, DATE(added) as DateNoTime 
FROM (
    select min(DateNoTime) as MinDate 
    from 
    (
     select DATE(added) as DateNoTime 
     FROM tb_videos 
     order by DateNoTime desc 
     limit 5 
    ) x) y, tb_videos 
WHERE added >= y.MinDate 
ORDER BY added desc 

这给你所有的数据。在PHP中，记录DateNoTime。无论何时发生变化，您都处于不同的日期，这会导致您触发另一个查询。否则，代码应该只是最低限度地改变。

未经检查的PHP代码

$result = mysql_query(' 
    SELECT *, DATE(added) as DateNoTime 
    FROM (
     select min(DateNoTime) as MinDate 
     from 
     (
      select DATE(added) as DateNoTime 
      FROM tb_videos 
      order by DateNoTime desc 
      limit 5 
     ) x) y, tb_videos 
    WHERE added >= y.MinDate 
    ORDER BY added desc 
'); 
$prevdate = NULL; 
foreach($result as $rst) 
{  
    if($prevdate!=$rst=>DateNoTime) { 
     // do something, like printing out a header for each new date 
     $prevdate=$rst=>DateNoTime; 
    } 

    // do something with the record 
}