SQlite 仅选择它们之间具有已定义差异的行

Question

例如我有以下行

'1982-01-10T00:00:00Z'
'1982-01-11T00:00:00Z'
'1982-01-14T00:00:00Z'
'1985-01-16T00:00:00Z'
'1985-01-17T00:00:00Z'
'1985-02-12T00:00:00Z'
'1987-01-11T00:00:00Z'
'1987-01-12T00:00:00Z'
'1987-01-13T00:00:00Z'

我只需要第一行，第一行和第二行之间的差异不超过 1 天，我也想获得具有这种差异的行数，对于这个示例，我想得到以下内容:

'1982-01-10T00:00:00Z', 2
'1985-01-16T00:00:00Z', 2
'1987-01-11T00:00:00Z', 3

有什么想法吗？

我尝试过查询，但结果错误:

SELECT utc_timestamp, utc_timestamp - LAG (utc_timestamp, 1, utc_timestamp) OVER (
        ORDER BY utc_timestamp
    ) difference
FROM (
    SELECT utc_timestamp, AVG(GB_temperature) as avgt
    FROM weather_data
    GROUP BY strftime('%Y-%m-%d', utc_timestamp)
    HAVING avgt < -4
);

Answer 1

嗯，这看起来不错，但我相信它可以用更少的代码行完成......

select min(date_before), count(date_c)+1, month, year from
(select strftime('%d',date_c) - lag(strftime('%d',date_c)) over (order by date_c) diff
       , strftime('%d',date_c) day
       , lag(strftime('%d', date_c)) over (order by date_c) day_before
       , strftime('%m', date_c) month
       ,  lag(strftime('%m', date_c)) over (order by date_c) m_before              
       , strftime('%Y', date_c) year
       , lag(strftime('%Y', date_c)) over (order by date_c) y_before
       , date_c
       , lag(date_c) over (order by date_c) date_before
from testTable
order by date_c)
where diff = 1
and month = m_before
and year = y_before
group by  month, year;

这是DEMO