Question

我有一个名为asset_usages的表，用于记录查看者对资产的查看。相关领域是

id (int)
asset_id (int)
viewer_type (string)
viewer_id (int)
viewed_at (datetime)

我刚刚添加了一个名为time_between_viewings的新字段，这是一个表示秒的int字段。我想将此设置为自上次查看该资产以来的时间（以秒为单位）。所以，如果我有这四个记录：

+-----+----------+-----------+-------------+---------------------+-----------------------+
| id  | asset_id | viewer_id | viewer_type | viewed_at           | time_between_viewings |
+-----+----------+-----------+-------------+---------------------+-----------------------+
| 506 |     7342 |      1182 | User        | 2009-01-05 11:10:01 |      NULL             |
| 509 |     7342 |      1182 | User        | 2009-01-05 11:12:47 |      NULL             |
| 514 |     6185 |      1182 | User        | 2009-01-05 11:14:28 |      NULL             |
| 524 |     6185 |      1182 | User        | 2009-01-05 11:28:18 |      NULL             |
| 618 |     1234 |      1182 | User        | 2009-01-05 11:29:03 |      NULL             |
| 729 |     1234 |      1182 | User        | 2009-01-05 11:29:01 |      NULL             |        
+-----+----------+-----------+-------------+---------------------+-----------------------+

然后time_between_viewings应设置如下：

+-----+----------+-----------+-------------+---------------------+-----------------------+
| id  | asset_id | viewer_id | viewer_type | viewed_at           | time_between_viewings |
+-----+----------+-----------+-------------+---------------------+-----------------------+
| 506 |     7342 |      1182 | User        | 2009-01-05 11:10:01 |      NULL             |
| 509 |     7342 |      1182 | User        | 2009-01-05 11:12:47 |      166              |
| 514 |     6185 |      1182 | User        | 2009-01-05 11:14:28 |      NULL             |
| 524 |     6185 |      1182 | User        | 2009-01-05 11:28:18 |      830              |
| 618 |     1234 |      1182 | User        | 2009-01-05 11:29:03 |      2                |
| 729 |     1234 |      1182 | User        | 2009-01-05 11:29:01 |      NULL             |     
+-----+----------+-----------+-------------+---------------------+-----------------------+

其中166和830是每对之间的时差，以秒为单位。

填充此字段的sql是什么？我无法弄明白。

重要说明：数据并非始终按时间顺序插入数据库。即，您可以有两个记录A和B，其中B具有更高的ID但A具有更高的lit_at值。因此，查找具有较低ID的第一个匹配记录不一定会让您先前查看同一个人 - 您需要检查数据库中的所有记录。

谢谢！最大

编辑 - 声明time_between_viewings是一个表示秒的int字段。

编辑 - 添加了几行作为具有更高内容但更早时间戳的行的示例

编辑 - 我刚刚意识到我没有正确地规定这个问题。 time_between_viewings应该等于同一个查看者上次查看资产以来的时间，即记录与具有相同asset_id，viewer_id和viewer_type的上一个（基于Viewed_at）记录之间的时间。我给出的示例数据仍然保留，但我可以放入一些不同的viewer_id和viewer_type值来稍微充实示例。

Answer 1

如果您准备了样本表和数据插页，将会有所帮助阅读此链接，了解为什么如果您想获得帮助，这一点非常重要：http://tkyte.blogspot.com/2005/06/how-to-ask-questions.html

这次我为您创建了它，点击此链接：http://sqlfiddle.com/#!2/9719a/2

并尝试此查询（您将在上述链接下找到此查询以及示例数据）：

select alias1.*,
       timestampdiff( second, previous_viewed_at, viewed_at ) 
         as time_between_viewings
from (
select alias.*,
       (
         select viewed_at from (
             select
             ( select count(*) from asset_usages y
                where x.asset_id = y.asset_id 
                and y.viewed_at < x.viewed_at
              ) as rn,
              x.* 
              from asset_usages x
          ) xyz
          where xyz.asset_id = alias.asset_id 
               and xyz.rn = alias.rn - 1
       ) previous_viewed_at
from (
  select
    ( select count(*) from asset_usages y
      where x.asset_id = y.asset_id 
        and y.viewed_at < x.viewed_at
    ) as rn,
    x.* 
  from asset_usages x
) alias
) alias1;

Answer 2

此SELECT语句将为您提供正确的数据。但是，您可能需要以块的形式进行更新。

您可以删除UPDATE语句的ORDER BY子句。如上所述，派生数据不依赖于外部SELECT语句中的行顺序。

select asset_id, viewer_id, viewer_type, viewed_at, 
       prev_viewed_at, timestampdiff(second, prev_viewed_at, viewed_at) elapsed_sec
from (select asset_id, viewer_id, viewer_type, viewed_at,
            (select max(t2.viewed_at) 
             from Table1 t2
             where t2.viewed_at < Table1.viewed_at
               and t2.asset_id = Table1.asset_id
               and t2.viewer_id = Table1.viewer_id
            ) prev_viewed_at
      from Table1
      )t3
order by asset_id, viewer_id, viewed_at;

存储记录之间的时间

2 个答案: