BigQuery，SQL：从两个时间戳中提取时间差异

Question

我有一张这样的桌子：

event_timestamp                               SESSION_ID    
2021-01-24 15:22:33.761011 UTC  0046aa0e-5035-4b2d-a6de-77a47949cdf6    
2021-01-24 15:22:33.761011 UTC  0046aa0e-5035-4b2d-a6de-77a47949cdf6    
2021-01-24 15:22:53.335013 UTC  0046aa0e-5035-4b2d-a6de-77a47949cdf6    
2021-01-24 15:24:53.700444 UTC  0046aa0e-5035-4b2d-a6de-77a47949cdf6    
2021-01-24 15:25:05.362517 UTC  0046aa0e-5035-4b2d-a6de-77a47949cdf6    
2021-01-24 15:25:05.362517 UTC  0046aa0e-5035-4b2d-a6de-77a47949cdf6    
2021-01-24 15:25:48.949679 UTC  0046aa0e-5035-4b2d-a6de-77a47949cdf6    
2021-01-24 15:25:48.949679 UTC  0046aa0e-5035-4b2d-a6de-77a47949cdf6    
2021-01-24 15:27:47.543331 UTC  0046aa0e-5035-4b2d-a6de-77a47949cdf6    
2021-01-24 15:27:47.543331 UTC  0046aa0e-5035-4b2d-a6de-77a47949cdf6    
2021-01-24 15:28:22.442896 UTC  0046aa0e-5035-4b2d-a6de-77a47949cdf6    
2021-01-24 15:28:22.442896 UTC  0046aa0e-5035-4b2d-a6de-77a47949cdf6    
2021-01-24 15:34:41.572305 UTC  0046aa0e-5035-4b2d-a6de-77a47949cdf6    
2021-01-24 16:37:28.988887 UTC  0046aa0e-5035-4b2d-a6de-77a47949cdf6    
2021-01-24 16:37:28.988887 UTC  0046aa0e-5035-4b2d-a6de-77a47949cdf6    

我需要计算平均 session 时间：

with cte as (
SELECT
timestamp_micros(event_timestamp) as timestamp,
min(timestamp_micros(event_timestamp)) over (partition by  SESSION_ID) as min_time,
max(timestamp_micros(event_timestamp)) over (partition by  SESSION_ID) as max_time,
SESSION_ID
FROM 'table'
)
select 
timestamp
round(avg (timestamp_diff(min_time,max_time,second)) over (partition by c.SESSION_ID)/60,2) as AVG_session_time,
SESSION_ID

Output：

event_timestamp                     AVG_session_time    SESSION_ID
2021-01-24 15:22:33.761011 UTC        74.92         0046aa0e-5035-4b2d-a6de-77a47949cdf6
2021-01-24 15:22:33.761011 UTC        74.92         0046aa0e-5035-4b2d-a6de-77a47949cdf6
2021-01-24 15:22:53.335013 UTC        74.92         0046aa0e-5035-4b2d-a6de-77a47949cdf6
2021-01-24 15:24:53.700444 UTC        74.92         0046aa0e-5035-4b2d-a6de-77a47949cdf6
2021-01-24 15:25:05.362517 UTC        74.92         0046aa0e-5035-4b2d-a6de-77a47949cdf6
2021-01-24 15:25:05.362517 UTC        74.92         0046aa0e-5035-4b2d-a6de-77a47949cdf6
2021-01-24 15:25:48.949679 UTC        74.92         0046aa0e-5035-4b2d-a6de-77a47949cdf6
2021-01-24 15:25:48.949679 UTC        74.92         0046aa0e-5035-4b2d-a6de-77a47949cdf6
2021-01-24 15:27:47.543331 UTC        74.92         0046aa0e-5035-4b2d-a6de-77a47949cdf6
2021-01-24 15:27:47.543331 UTC        74.92         0046aa0e-5035-4b2d-a6de-77a47949cdf6
2021-01-24 15:28:22.442896 UTC        74.92         0046aa0e-5035-4b2d-a6de-77a47949cdf6
2021-01-24 15:28:22.442896 UTC        74.92         0046aa0e-5035-4b2d-a6de-77a47949cdf6
2021-01-24 15:34:41.572305 UTC        74.92         0046aa0e-5035-4b2d-a6de-77a47949cdf6
2021-01-24 16:37:28.988887 UTC        74.92         0046aa0e-5035-4b2d-a6de-77a47949cdf6
2021-01-24 16:37:28.988887 UTC        74.92         0046aa0e-5035-4b2d-a6de-77a47949cdf6

但是 74.92 不是时差，因为这 74 分钟和 60 秒的 92% 所以我需要得到像 0.92*60/100 = 74:55 这样的答案 我无法提取秒的百分比并计算秒，因为有文本和数字之间的转换和提取的问题......有没有另一种方法可以得到时间（分和秒）形式的结果？？？

谢谢 ）

Answer 1

考虑下面

select average_session_duration,
    format(
        '%i day %i hour %i minute %i second', 
        unix_date(date(timestamp_seconds(average_session_duration))), 
        extract(hour from time(timestamp_seconds(average_session_duration))), 
        extract(minute from time(timestamp_seconds(average_session_duration))), 
        extract(second from time(timestamp_seconds(average_session_duration)))
    ) as average_duration_as_long_string,
    regexp_replace(
        cast(time(timestamp_seconds(average_session_duration)) as string), 
        r'^\d\d', 
        cast(extract(hour from time(timestamp_seconds(average_session_duration))) + 24 * unix_date(date(timestamp_seconds(average_session_duration))) as string)
    ) as average_duration_as_short_string
from (
    select cast(avg(session_duration) as int64) as average_session_duration from (
        select 
            timestamp_diff(max(event_timestamp), min(event_timestamp), second) session_duration
        from `project.dataset.table`
        group by SESSION_ID
    )
)

如果适用于您的问题中的示例数据 - output 是