Question

我试图找到在给定时间段内承诺最少的所有LID。以下是我的表：

| requestDate |    BID    |    LID    | PercentageContributed | DateComitted
| 1/10/2016   | S7444319C | S2262450A |           0.25        |   18/10/2016
| 1/10/2016   | S7444319C | S4495282I |           0.25        |   7/10/2016
| 1/10/2016   | S7444319C | S4792394D |           0.25        |   28/10/2016
| 1/10/2016   | S7444319C | S7173102K |           0.25        |   9/10/2016
| 25/10/2016  | S9981233W | S2848191X |           1.00        |   13/11/2016
| 28/10/2016  | S7611209X | S2848191X |           0.33        |   13/11/2016
| 28/10/2016  | S7611209X | S4792394D |           0.33        |   4/11/2016
| 28/10/2016  | S7611209X | S7145303Q |           0.33        |   5/11/2016

我尝试过以下代码：

SELECT LID, COUNT(*) AS TotalNumCommitte
FROM Commits
WHERE DateCommitted BETWEEN '2015-11-03' AND '2017-01-09'
GROUP BY LID

但我只是设法得到以下内容：

|    LID    | TotalNumCommitte
| S2262450A |        1
| S2848191X |        2
| S4495282I |        1
| S4792394D |        2
| S7145303Q |        1
| S7173102K |        1

我想要实现的目标如下：

|    LID    | TotalNumCommitte
| S2262450A |        1
| S4495282I |        1
| S7145303Q |        1
| S7173102K |        1

我也尝试了以下内容：

SELECT MIN(TotalNumCommitted)
FROM (
    SELECT LID, COUNT(*) AS TotalNumCommitted
    FROM Commits
    WHERE DateCommitted >= '2015-11-03'
    AND DateCommitted <= '2017-01-09'
    GROUP BY LID
) T1

但我只得到1（这是最低限度）非常感谢你的帮助。

Answer 1

假设现代SQL支持公用表表达式和排名函数，如下所示：

;WITH Counts as (
    SELECT LID, COUNT(*) AS TotalNumCommitte
    FROM Commits
    WHERE DateCommitted BETWEEN '2015-11-03' AND '2017-01-09'
    GROUP BY LID
), RankedCounts as (
    SELECT *,RANK() OVER (ORDER BY TotalNumCommitte) as rk
    FROM Counts
)
select *
from RankedCounts
where rk=1

Answer 2

我认为最简单的方法是TOP WITH TIES：

SELECT TOP (1) WITH TIES LID, COUNT(*) AS TotalNumCommitte
FROM Commits
WHERE DateCommitted BETWEEN '2015-11-03' AND '2017-01-09'
GROUP BY LID
ORDER BY TotalNumCommitte;

Answer 3

在标准SQL中，您使用{/ 1}}，如

FETCH FIRST ROW WITH TIES

在SQL Server中，您必须改为使用SELECT LID, COUNT(*) AS TotalNumCommitte FROM Commits WHERE DateCommitted BETWEEN '2015-11-03' AND '2017-01-09' GROUP BY LID ORDER BY COUNT(*) FETCH FIRST ROW WITH TIES;：

TOP (1) WITH TIES

Answer 4

您可以将查询组与窗口函数组合使用：

select lid, totalnumcommitte
from (
    select lid, 
           count(*) as totalnumcommitte, 
           rank() over (order by count(*)) as rnk
    from commits
    group by lid
) t
where rnk = 1
order by lid;

或使用公用表表达式：

with counts as (
    select lid, 
           count(*) as totalnumcommitte, 
           rank() over (order by count(*)) as rnk
    from commits
    group by lid
)
select lid, totalnumcommitte
from counts
where rnk = 1
order by lid;

Answer 5

SELECT LID, COUNT(*) AS TotalNumCommitte FROM Commits WHERE DateCommitted BETWEEN '2015-11-03' AND '2017-01-09'
GROUP BY LID 
HAVING COUNT(*) =
(SELECT TOP 1 COUNT(*) AS TotalNumCommitted FROM Commits WHERE DateCommitted BETWEEN '2015-11-03' AND '2017-01-09'
GROUP BY LID ORDER BY TotalNumCommitted)

Answer 6

您可以使用HAVING过滤聚合函数的结果（在本例中为计数），例如：

SELECT LID, COUNT(*) AS TotalNumCommitte
FROM Commits
WHERE DateCommitted BETWEEN '2015-11-03' AND '2017-01-09'
GROUP BY LID HAVING COUNT(*) = 1

在Count中找到最少（*）

6 个答案: