计算 24 小时内的请求数以确定用户会话?



>想象一个包含以下行的Web服务器日志:

<timestamp> <ip> <user-agent> <product page>

我想要一份报告

  • 使用以下条件计算 24 小时内每个用户会话对产品页面的请求数:
  • 唯一用户定义为多个列的组合 ( (
  • 24小时
  • 窗口从第一次请求产品页面的时间戳开始(24小时窗口可以在任何时间开始(
  • 如果请求时间戳之间相隔 24 小时,则将被视为新用户会话

对于以下日志:

2019-1-1 01:00 1.2.3.4 Netscape product 5
2019-1-1 01:01 1.2.3.4 Netscape product 5
2019-1-1 01:00 1.2.3.5 Chrome product 5
2019-1-1 01:01 1.2.3.5 Chrome product 5
2019-1-1 01:59 1.2.3.4 Netscape product 5
2019-1-1 02:00 1.2.3.4 Netscape product 4
2019-1-1 02:01 1.2.3.4 Netscape product 4
2019-1-1 02:02 1.2.3.4 Netscape product 4
2019-1-1 07:43 1.2.3.5 Chrome product 5
2019-1-2  2:01 1.2.3.4 Netscape product 5

将产生:

1.2.3.4/Netscape, product 4, 1
1.2.3.4/Netscape, product 5, 2
1.2.3.5/Chrome: product 5, 1

也许第二个查询会输出:

1.2.3.4/Netscape, 6
1.2.3.4/Netscape, 1
1.2.3.5/Chrome, 3

(每个用户 24 小时窗口的请求数,因此 1.2.3.4/Netscape 列出了两次(

将提供上述两个结果集的示例查询是什么?

奖励/可选:如果 24 小时内的请求间隔超过 30m,将被视为另一个新会话

这里有一些东西可以给你一个方向(虽然不一定太高性能/效率,这取决于输入数据集的大小(。

datatable(timestamp:datetime, ip:string, user_agent:string, product_page:string)
[
datetime(2019-01-01 01:00), '1.2.3.4', 'Netscape', 'product 5',
datetime(2019-01-01 01:01), '1.2.3.4', 'Netscape', 'product 5',
datetime(2019-01-01 01:00), '1.2.3.5', 'Chrome',   'product 5',
datetime(2019-01-01 01:01), '1.2.3.5', 'Chrome',   'product 5',
datetime(2019-01-01 01:59), '1.2.3.4', 'Netscape', 'product 5',
datetime(2019-01-01 02:00), '1.2.3.4', 'Netscape', 'product 4',
datetime(2019-01-01 02:01), '1.2.3.4', 'Netscape', 'product 4',
datetime(2019-01-01 02:02), '1.2.3.4', 'Netscape', 'product 4',
datetime(2019-01-01 07:43), '1.2.3.5', 'Chrome',   'product 5',
datetime(2019-01-02 02:01), '1.2.3.4', 'Netscape', 'product 5',
]
| extend user = strcat(ip, "/", user_agent)
| order by user asc, timestamp asc
| extend session_start = row_window_session(timestamp, 24h, 24h, user_agent != prev(user_agent) or product_page != prev(product_page) or ip != prev(ip))
| summarize session_count = dcount(session_start) by user, product_page

-->

| user             | product_page | session_count |
|------------------|--------------|---------------|
| 1.2.3.4/Netscape | product 5    | 2             |
| 1.2.3.4/Netscape | product 4    | 1             |
| 1.2.3.5/Chrome   | product 5    | 1             |

对于第二个查询,以下内容可能有效:

datatable(timestamp:datetime, ip:string, user_agent:string, product_page:string)
[
datetime(2019-01-01 01:00), '1.2.3.4', 'Netscape', 'product 5',
datetime(2019-01-01 01:01), '1.2.3.4', 'Netscape', 'product 5',
datetime(2019-01-01 01:00), '1.2.3.5', 'Chrome',   'product 5',
datetime(2019-01-01 01:01), '1.2.3.5', 'Chrome',   'product 5',
datetime(2019-01-01 01:59), '1.2.3.4', 'Netscape', 'product 5',
datetime(2019-01-01 02:00), '1.2.3.4', 'Netscape', 'product 4',
datetime(2019-01-01 02:01), '1.2.3.4', 'Netscape', 'product 4',
datetime(2019-01-01 02:02), '1.2.3.4', 'Netscape', 'product 4',
datetime(2019-01-01 07:43), '1.2.3.5', 'Chrome',   'product 5',
datetime(2019-01-02 02:01), '1.2.3.4', 'Netscape', 'product 5',
]
| extend user = strcat(ip, "/", user_agent)
| summarize count() by user, startofday(timestamp)
| project-away timestamp

-->

| user             | count_ |
|------------------|--------|
| 1.2.3.4/Netscape | 6      |
| 1.2.3.5/Chrome   | 3      |
| 1.2.3.4/Netscape | 1      |

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