回答編集履歴
2
推敲
test
CHANGED
|
@@ -1,3 +1,5 @@
|
|
|
1
|
+
1,3を1 以外を9に置き換え、加算した結果が２になるものを抽出
|
|
2
|
+
|
|
1
3
|
```SQL
|
|
2
4
|
|
|
3
5
|
SELECT *
|
1
訂正
test
CHANGED
|
@@ -12,40 +12,10 @@
|
|
|
12
12
|
|
|
13
13
|
from address_type
|
|
14
14
|
|
|
15
|
-
where type_id in (1,3)
|
|
16
|
-
|
|
17
15
|
group by address_id
|
|
18
16
|
|
|
19
|
-
having c
|
|
17
|
+
having sum(case when type_id in (1,3) then 1 else 9 end)=2
|
|
20
18
|
|
|
21
19
|
)
|
|
22
20
|
|
|
23
21
|
```
|
|
24
|
-
|
|
25
|
-
name に関する条件による件数が少ないなら以下の方が、高速かもしれません。
|
|
26
|
-
|
|
27
|
-
```SQL
|
|
28
|
-
|
|
29
|
-
SELECT *
|
|
30
|
-
|
|
31
|
-
FROM address_type t left outer join address a on t.address_id = a.id
|
|
32
|
-
|
|
33
|
-
WHERE a.name IN ('a1@d1','a4@d1','a5@d1','a6@d1')
|
|
34
|
-
|
|
35
|
-
AND exists (
|
|
36
|
-
|
|
37
|
-
select 1 from address_type
|
|
38
|
-
|
|
39
|
-
where type_id = 1 and address_id=t.address_id
|
|
40
|
-
|
|
41
|
-
)
|
|
42
|
-
|
|
43
|
-
AND exists (
|
|
44
|
-
|
|
45
|
-
select 1 from address_type
|
|
46
|
-
|
|
47
|
-
where type_id = 3 and address_id=t.address_id
|
|
48
|
-
|
|
49
|
-
)
|
|
50
|
-
|
|
51
|
-
```
|