Question

我正在尝试将用户初始日期的时间（小时分和秒）添加到当前日期，但收到错误。

我不知道我做错了什么，对其他建议持开放态度

select distinct uid,
       min (ts) over (partition by uid order by ts rows unbounded preceding) as initial_date,
       current_date-2 + interval 'to_char(min (ts) over (partition by uid order by ts rows unbounded preceding),'HH') hours'
  from mr_session_log
where ts >= '2015-09-01'

这就是我收到的错误

ERROR: syntax error at or near "') hours'"

职位：355

current_date-2 + interval 'to_char(min (ts) over (partition by uid order by ts rows unbounded preceding),'HH') hours',
                                                                                                               ^

Answer 1

你不应该过度设计解决方案，语法高亮显示你不能把函数放在间隔中，当你想要uid和uid分区时，不需要窗口函数。

SELECT uid, (current_date -2 ) + min(ts::time)
FROM mr_session_log
WHERE ts >= '2015-09-01'
GROUP BY uid
ORDER BY uid;

Answer 2

如果你真的需要使用窗口功能，它也很简单：

SELECT DISTINCT
    uid,
    CURRENT_DATE - 2 - MIN(ts::time) OVER (PARTITION BY uid
                                     ORDER BY ts
                                     ROWS UNBOUNDED PRECEDING) AS date_with_added_time
FROM mr_session_log
WHERE ts >= '2015-09-01'::DATE;

如果您只需要小时，分钟和秒，请使用MIN(date_trunc('secconds', ts)::time)代替MIN(ts::time)

将时间从一个时间戳字段添加到其他字段

2 个答案: