Question

我有一个用于记录方法调用的表。它有LogId，MethodId，DateTime列。

我需要编写一个select语句，该语句计算特定时间段内特定方法ID的所有日志，并显示特定方法在不同时间段内的日志数。

第一点很简单：

select
    l.[MethodId],
    count(l.[LogId]) as [Count]
from
    [Log] l (nolock)
where
    l.[DateTime] between @from and @to
    and l.[MethodId] in @methodIds
group by
    l.[MethodId]

但是现在我需要该表中的第二列，Previous，如果它在一个单独的语句中，它将如下所示：

select
    l.[MethodId],
    count(l.[LogId]) as [Previous]
from
    [Log] l (nolock)
where
    l.[DateTime] between @before and @from 
    and l.[MethodId] in @methodIds
group by
    l.[MethodId]

并非所有方法都会有两个时间段的日志，因此如果连接在这些情况下在count / previous列中插入0而不是它们为null，那将会很好。如果某个方法在任何时段都没有任何日志，那就没问题。

我希望在一个表格中看到MethodId, Count, Previous。我该如何实现这一目标？

Answer 1

类似的东西：

select 
    l.[MethodId], 
    sum(case when datetime between @from and @to then 1 else 0 end) as count,
    sum(case when datetime between @before and @from then 1 else 0 end) as previous
from 
    [Log] l
where 
    l.[DateTime] between @before and @to 
    and l.[MethodId] in @methodIds 
group by 
    l.[MethodId]

where中的BETWEEN子句不会影响输出，但如果你有一个datetime索引，它可能会影响性能。如果这个表变大，你可能应该有这样的索引。

Answer 2

试试这个：

select
    l.[MethodId],
    count(isnull(l.[LogId],0)) as [Previous]
from
    [Log] l (nolock)
where
    l.[DateTime] between @before and @from 
    and l.[MethodId] in @methodIds
group by
    l.[MethodId]

Answer 3

尝试使用完全外连接：

DECLARE @from1 DATETIME, 
    @from2 DATETIME,
    @to1 DATETIME,
    @to2 DATETIME;

SELECT @from1 = '2011-01-01T00:00:00.000', 
    @to1 = '2011-01-31T23:59:59.997',
    @from2 = '2011-02-01T00:00:00.000',
    @to2 = '2011-02-28T23:59:59.997';


SELECT ISNULL(a.MethodID, b.MethodID) MethodID
    ,ISNULL(a.[Count]) CountA
    ,ISNULL(b.[Count]) CountB
FROM
(
select
    l.[MethodId],
    count(l.[LogId]) as [Count]
from
    [Log] l (nolock)
where
    l.[DateTime] between @from1 and @to1
    and l.[MethodId] in @methodIds
group by
    l.[MethodId]
) a 
FULL OUTER JOIN
(
select
    l.[MethodId],
    count(l.[LogId]) as [Previous]
from
    [Log] l (nolock)
where
    l.[DateTime] between @from2 and @to2 
    and l.[MethodId] in @methodIds
group by
    l.[MethodId]
) b
ON a.MethodID = b.MethodID

两组中的值由select语句合二为一

3 个答案: