我如何找到差距在sqlite表

Question

我有一个以毫秒为主键的时间戳的sqlite表，每一行之间应该间隔1秒或1000秒。有时我的数据记录仪坏了，表中没有那个时间的数据。我如何使用SQL语句找到差距?我知道基于游标的解决方案是可能的。

table = PVT
TS
1119636081000
1119636082000
1119636083000
1119636084000
1119636085000
------gap------
1119636090000
1119636091000

Answer 1

这可能有效。假设表名为"tstamps"，

select a.ts
from tstamps a
where not exists
   (select b.ts
    from tstamps b
    where b.ts = a.ts+1000)
and exists
   (select c.ts
    from tstamps c
    where c.ts = a.ts+2000)

另一种方式

select a.ts
from tstamps a
where not exists
   (select b.ts
    from tstamps b
    where b.ts = a.ts+1000)
and a.ts <
   (select max(c.ts)
    from tstamps c
   )

使用减号运算符。我不确定，这些查询哪一个性能更好。

select ts+1000
from pvt
where ts != (select max(ts) from pvt)
minus
select ts
from pvt
where ts != (select min(ts) from pvt)

Answer 2

类似这样(假设PVT.TS是您的列名):

SELECT * FROM 'table' WHERE PVT.TS ISNULL;

或

SELECT * FROM 'table' WHERE PVT.TS IS NULL;

如果您的收集器实际上正在输入一个空白条目，您可能需要

WHERE PVT.TS = ''

或

where ifnull(some_column, '') = ''

Answer 3

在写这篇文章的时候，SQLite不支持像LAG(TS) OVER (ORDER BY TS ASC)和LEAD() OVER这样的窗口函数，它们很容易分别给出前面和后面的TS值。

所以，你需要自己做:

sqlite> .mode col
sqlite> .width 14 14 14
sqlite>    SELECT PVT.TS  AS measurement,
                  prev.TS AS prev,
                  next.TS AS next
             FROM PVT
        LEFT JOIN PVT next ON PVT.TS = (next.TS - 1000)
        LEFT JOIN PVT prev ON PVT.TS = (prev.TS + 1000);

这将给你类似这样的东西(我使用不同的数据，你会看到):

-- measurement        prev            next     
  -------------   -------------   -------------
  1119636081000                   1119636082000   -- gap (no previous at all)
  1119636082000   1119636081000   1119636083000 
  1119636083000   1119636082000   1119636084000 
  1119636084000   1119636083000   1119636085000 
  1119636085000   1119636084000                   -- gap (no next offset 1000)
  1119636088000                   1119636089000   -- gap (no previous offset 1000)
  1119636089000   1119636088000                   -- gap (no next at all)

您始终可以将该查询限制为仅针对WHERE prev.TS IS NULL OR next.TS is NULL的记录。

Answer 4

创建一个至少包含86400行(每天每秒一行)的统计表:

create table Tally(n integer primary key not null);
insert into Tally(n) values (0),(1),(2),(3),(4),(5),(6),(7),(8),(9);
insert into Tally(n) select null from tally n1 , tally n2, tally n3, tally n4, tally n5;

将您的PVT表连接到当天的转置记录:

select 1119636081000 + tally.n*1000 as Expected, pvt.ts from tally left join pvt on pvt.ts = 1119636081000 + tally.n*1000 where tally.n <= 86400 limit 15;

给定一个表，我用您的样本数据填充，我得到这个作为输出:

Expected       TS
-------------  -------------
1119636081000  1119636081000
1119636082000  1119636082000
1119636083000  1119636083000
1119636084000  1119636084000
1119636085000  1119636085000
1119636086000
1119636087000
1119636088000
1119636089000
1119636090000  1119636090000
1119636091000  1119636091000
1119636092000
1119636093000
1119636094000
1119636095000

如果你在PVT.TS为空的地方进行额外的过滤，你应该得到缺失的值:

select 1119636081000 + tally.n*1000 as Expected, pvt.ts from tally left join pvt on pvt.ts = 1119636081000 + tally.n*1000 where tally.n <= 86400 and PVT.ts is null limit 15;
Expected       TS
-------------  ----------
1119636086000
1119636087000
1119636088000
1119636089000
1119636092000
1119636093000
1119636094000
1119636095000
1119636096000
1119636097000
1119636098000
1119636099000
1119636100000
1119636101000
1119636102000

注意:我使用限制15来保持我在控制台的理智