Question

时

我发现我的病情解决方案

COUNT(total) = 1不起作用：

SELECT H.hacker_id AS hacker_id, H.name AS name, COUNT(C.challenge_id) AS total
  FROM Hackers H
    JOIN Challenges C ON H.hacker_id = C.hacker_id
  GROUP BY H.hacker_id, H.name
    HAVING
      COUNT(total) = 1 OR -- HERE
      OR
      total = (
        SELECT MAX(amount) FROM (
          SELECT COUNT(challenge_id) AS amount FROM Challenges
          GROUP BY hacker_id
        ) t
      )
  ORDER BY total DESC, hacker_id;

输出：

5120 Julia 50 
18425 Anna 50 
20023 Brian 50 
33625 Jason 50 
41805 Benjamin 50 
52462 Nicholas 50 
64036 Craig 50 
69471 Michelle 50 
77173 Mildred 50 
94278 Dennis 50 
96009 Russell 50 
96716 Emily 50 
2689 Joe 1 
3432 Linda 1 
5827 Kelly 1 
6011 Robin 1 
7537 David 1 
7961 Michelle 1 
9642 Joseph 1 
9901 Lawrence 1 
10975 Christine 1 
11715 Louise 1 
13075 John 1 
13905 Evelyn 1 
14307 David 1 
14726 Emily 1 
15109 Dorothy 1 
17282 Norma 1 
17311 Andrew 1 
17663 Benjamin 1 
17846 Alan 1 
18709 James 1 
19032 Andrew 1 
19781 Jesse 1 
19962 Patricia 1 
20190 Aaron 1 
21350 Bobby 1 
21821 Carol 1 
21916 Clarence 1 
22396 Wayne 1 
22455 Carolyn 1 
23812 Jerry 1 
24047 Elizabeth 1 
25684 Alan 1 
25990 Mildred 1 
26802 Paul 1 
27797 Helen 1 
28766 Paul 1 
29242 Jennifer 1 
29841 Charles 1 
30677 Keith 1 
30778 Jose 1 
30908 Helen 1 
31770 Ashley 1 
32364 Julia 1 
32735 Cheryl 1 
33273 Sara 1 
33489 Denise 1 
37092 Jennifer 1 
37764 Jimmy 1 
38540 Katherine 1 
42467 Ernest 1 
43240 Diana 1 
43398 Steve 1 
43595 Adam 1 
45685 Bobby 1 
46144 Sharon 1 
46468 Timothy 1 
46604 Christina 1 
47156 Kelly 1 
47921 Lillian 1 
50560 Brian 1 
52350 Teresa 1 
53451 Jeffrey 1 
53597 Rose 1 
53768 Douglas 1 
54015 Carolyn 1 
54510 Paula 1 
55415 Amy 1 
56039 Teresa 1 
56103 Kelly 1 
56338 Jose 1 
57195 Beverly 1 
57873 Diana 1 
58086 Debra 1 
58167 David 1 
58543 Rachel 1 
59871 Martin 1 
59895 Martha 1 
60177 Brian 1 
61093 Mark 1 
61102 Kenneth 1 
61206 Lillian 1 
61769 Marie 1 
63263 Dorothy 1 
63684 Randy 1 
63730 Sarah 1 
63803 Carolyn 1 
63961 Anna 1 
64341 Virginia 1 
64882 Roy 1 
68178 Gloria 1 
70499 Dennis 1 
72321 Julie 1 
72763 Julie 1 
73267 Jeremy 1 
73676 Linda 1 
74320 Pamela 1 
78615 Kathryn 1 
79612 Tina 1 
81652 Albert 1 
83308 Roy 1 
84739 Alan 1 
84938 Judy 1 
85094 Matthew 1 
86142 Douglas 1 
87040 Craig 1 
87885 Gregory 1 
88069 Jean 1 
88083 Anna 1 
88084 Alan 1 
88858 Bruce 1 
89514 Jeffrey 1 
89903 Katherine 1 
90276 Joyce 1 
90369 Christina 1 
91620 Debra 1 
92239 Shirley 1 
92920 Louis 1 
94337 Lillian 1 
94676 Patrick 1 
94746 Adam 1 
96521 Christine 1 
96773 Angela 1 
97338 Amy 1 
98785 Rose 1 
99101 Timothy 1 
99165 Nancy 1

但这可行：

SELECT H.hacker_id AS hacker_id, H.name AS name, COUNT(C.challenge_id) AS total
  FROM Hackers H
    JOIN Challenges C ON H.hacker_id = C.hacker_id
  GROUP BY H.hacker_id, H.name
    HAVING
      --
      total IN
        (SELECT t0.total
          FROM
            (SELECT count(*) AS total
            FROM challenges
            GROUP BY hacker_id) t0
          GROUP BY t0.total
        HAVING count(t0.total) = 1)
      -- instead of 'COUNT(total) = 1'
      OR
      total = (
        SELECT MAX(amount) FROM (
          SELECT COUNT(challenge_id) AS amount FROM Challenges
          GROUP BY hacker_id
        ) t
      )
  ORDER BY total DESC, hacker_id;

输出：

5120 Julia 50 
18425 Anna 50 
20023 Brian 50 
33625 Jason 50 
41805 Benjamin 50 
52462 Nicholas 50 
64036 Craig 50 
69471 Michelle 50 
77173 Mildred 50 
94278 Dennis 50 
96009 Russell 50 
96716 Emily 50 
72866 Eugene 42 
37068 Patrick 41 
12766 Jacqueline 40 
86280 Beverly 37 
19835 Joyce 36 
38316 Walter 35 
29483 Jeffrey 34 
23428 Arthur 33 
95437 George 32 
46963 Barbara 31 
87524 Norma 30 
84085 Johnny 29 
39582 Maria 28 
65843 Thomas 27 
5443 Paul 26 
52965 Bobby 25 
77105 Diana 24 
33787 Susan 23 
45855 Clarence 22 
33177 Jane 21 
7302 Victor 20 
54461 Janet 19 
42277 Sara 18 
99388 Mary 16 
31426 Carlos 15 
95010 Victor 14 
27071 Gerald 10 
90267 Edward 9 
72609 Bobby 8

所以，为什么我不能只用这个

COUNT(total) = 1

代替这种大条件：

total IN
  (SELECT t0.total
   FROM
     (SELECT count(*) AS total
      FROM challenges
      GROUP BY hacker_id) t0
   GROUP BY t0.total
   HAVING count(t0.total) = 1)

Answer 1

因为在您的上下文中，总计只是汇总结果的别名，而不是值。通常，您必须重复COUNT(C.challenge_id)部分，但这对您没有帮助，因为COUNT(COUNT(C.challenge_id)) = 1显然是错误的。

有一些解决方案，例如将聚合数据集结果放入临时表/表变量或使用CTE，这就是子查询起作用的原因。

免责声明：不要检查查询的正确性，但IN听起来很丑，使用CTE可能是更好的方法。

下面是使用MS SQL的示例解决方案，对不起，现在不要运行MySql，不要使用CTE =）

create table dbo.HACKER
(
    hacker_id int,
    name varchar(100)
)
GO

create table dbo.CHALLENGE
(
    challenge_id int,
    hacker_id int
)

GO

insert into dbo.HACKER
(hacker_id,name)
values
 (5077,'Rose')
,(21283,'Angela')
,(62743,'Frank')
,(88255,'Patrick')
,(96196,'Lisa')

insert into dbo.CHALLENGE
(challenge_id, hacker_id)
values
 (61654,5077)
,(58302,21283)

GO
--drop table #Temp
--drop table #totalsToExclude
select hk.hacker_id, hk.name, x.total
into #Temp
from dbo.HACKER hk
join (select ch.hacker_id, count(*) as total from dbo.CHALLENGE ch group by ch.hacker_id) as x
on x.hacker_id = hk.hacker_id

select * from #Temp

declare @maxTotal as int =(select max(total) from #temp)
select @maxTotal

select t.total, count(*) as [Count_total]
into #totalsToExclude
from #temp t
group by t.total
having(count(*) >1)

delete tx from #totalsToExclude tx where tx.total = @maxTotal
select * from #totalsToExclude

select * from #Temp t
where t.total not in (select t.total from #totalsToExclude)
order by t.total desc, t.hacker_id

为什么COUNT在SQL HAVING条件下无法计算正确的值？

1 个答案: