Question

我有一个表格,其结构如下:

id           -int(11)
event_id     -int(11)
photo_id     -int(11)
created_at   -datetime

我如何写一个回最近100个牢房的问询,但保证光电灯的连续浏览量不超过4个。

Answer 1

您可以添加一条规定,将4个排位排出,低于photo_id。存在:

select *
from YourTable t1
where 4 > (
    select count(*)
    from YourTable t2
    where t1.event_id = t2.event_id
    and t1.photo_id < t2.photo_id
)
limit 100

这对庞大的表格来说可能会有某种缓慢。一个更快但非常具体的MySQL方案是使用变量。例如:

select *
from (
    select
        @nr := case 
            when event_id = @event then @nr + 1 
            else 1 
        end as photonr
    ,   @event := event_id
    ,   t1.*
    from YourTable as t1
    cross join (select @event := -1, @nr := 1) as initvars
    order by event_id
) as subquery
where subquery.photonr < 5
limit 100;

使用的测试数据:

drop table if exists YourTable;

create table YourTable (
  id int auto_increment primary key
, event_id int
, photo_id int
);

insert into YourTable (event_id, photo_id)
values (1,1), (1,2), (1,3), (1,4), (1,5), (2,1), (1,6);

Answer 2

在矿石中,你将使用滞后功能。

LAG  (value_expression [,offset] [,default]) OVER ([query_partition_clause] order_by_clause)

我的后勤部无法肯定这一点。

Answer 3

如果您使用T-SQL,请查看用于排名功能。

从你的问题来看,它像NTILE是你想要的。在我迅速试图盘问时,我不停在终点站,以免检查,而是应该开始:

SELECT
  id,
  event_id,
  photo_id,
  created_at,
  NTILE(4) OVER (ORDER BY photo_id) AS  Quartile 
FROM tbl
WHERE NTILE(4) OVER (ORDER BY photo_id)<2
ORDER BY created_at DESC

链接网页有良好的例子说明所有排名的职能。

uck

Answer 4

为此:

SELECT p.id, p.event_id, p.photo_id, p.created_at
FROM photo_table p,
    (

        SELECT photo_id, MAX(created_at) max_date
        FROM photo_table
        GROUP BY photo_id 
    ) t
WHERE p.created_at = t.max_date
        AND p.photo_id = t.photo_id
ORDER BY p.created_at DESC
LIMIT 100

What it does is: 1. find latest photo change date 2. find only last events of each photo 3. select first 100 most recent

在PogreSQL或Oracle,使用Aalytica/windowing职能,如:

FIRST (created_at) OVER (PARTITION BY photo_id ORDER BY created_at DESC)

Answer 5

我这样说,你将走上正轨:

$sql = "SELECT DISTINCT * FROM myTable ORDER BY id ASC LIMIT 100";

在本案中,“DISTINCT”将只收回分流,无视重复的分行。

希望会有所助益。

友情链接