Sqlite - 如果有相同的号码,我如何像通话记录一样查询继续获取单个记录,仅连续记录计数?

Sqlite - How to i query like call log if there is same number continue get as single record with continuously record count only?

UniqueID | MobileNumber | createDate
-----------+---------+-----+------------+-----------
U_23121  | 987654        | 2013-02-05 
U_23124  | 987654        | 2013-02-02 
U_23122  | 845263        | 2013-01-18 
U_23128  | 654789        | 2013-01-16 
U_23123  | 735689        | 2013-01-12 
U_23128  | 654789        | 2013-01-11 
U_23128  | 654789        | 2013-01-10 
U_23126  | 987654        | 2013-01-09 
U_23125  | 845263        | 2013-01-07 
U_23126  | 845263        | 2013-01-06 
U_23125  | 987654        | 2013-01-05 

我想记录喜欢用手机号码过滤如果超过一个继续根据 createdDate 获取最新并获得计数

        UniqueID | Mobile_Number | createDate   | count
        -----------+---------+-----+------------+-----------
        U_23121  | 987654        | 2013-02-05   | 2
        U_23122  | 845263        | 2013-01-18   | 1
        U_23128  | 654789        | 2013-01-16   | 1
        U_23123  | 735689        | 2013-01-12   | 1
        U_23128  | 654789        | 2013-01-11   | 2
        U_23126  | 987654        | 2013-01-09   | 1
        U_23125  | 845263        | 2013-01-07   | 2
        U_23125  | 987654        | 2013-01-05   | 1

我将从以下查询中获取需要的记录但未获取计数

SELECT te.*
FROM tableName as te
WHERE te.Mobile_Number <> (select Mobile_Number 
                           from tableName
                           where createDate > te.createDate
                           limit 1
                          )
ORDER BY te.createDate DESC

使用分组依据

SELECT te.* FROM tableName as te where te.Mobile_Number != (select Mobile_Number from tableName where createDate > te.createDate limit 1) GROUP BY Mobile_Number  ORDER BY te.createDate DESC

好的,这就是间隙和孤岛问题。如果您的 Sqlite 支持 row_number 函数(版本 3.25 及更高版本),那么您可以使用以下方法

select MobileNumber, max(createDate), count(*)
from
(
  select *,
       row_number() over (order by createDate) -
       row_number() over (partition by MobileNumber order by createDate) grp
  from data
) t
group by grp, MobileNumber

这是一个缺口和孤岛问题。一种解决方案是为每一行分配一个 "grp",然后按该组聚合。

您可以通过计算每行中与手机号码不同 的手机号码的数量来分配 grp,直到该行。这是相邻手机号码的固定值。

结果查询:

SELECT MAX(UniqueId), MobileNumber,
       MAX(createDate), COUNT(*)
FROM (SELECT te.*,
             (SELECT COUNT(*)
              FROM tableName te2
              WHERE te2.createDate < te.createDate AND
                    te2.MobileNumber <> te.MobileNumber
             ) as grp
      FROM tableName te
     ) te
GROUP BY MobileNumber, grp;
ORDER BY MIN(tcreateDate) DESC