回答編集履歴
3
追記
test
CHANGED
@@ -6,7 +6,7 @@
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6
6
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7
7
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from (
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8
8
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9
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-
select AAA,CCC,(select count(*) from テーブル where BBB=t1.BBB) cnt
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9
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+
select AAA,CCC,(select count(*) from テーブル where AAA=t1.AAA and BBB=t1.BBB) cnt
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10
10
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11
11
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from テーブル t1
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12
12
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@@ -26,10 +26,32 @@
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26
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```SQL
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28
28
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29
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-
select AAA,CCC,(select count(*) from テーブル where BBB=t1.BBB) cnt
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+
select AAA,CCC,(select count(*) from テーブル where AAA=t1.AAA and BBB=t1.BBB) cnt
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30
30
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31
31
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from テーブル t1
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32
32
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33
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where CCC is not null
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34
34
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35
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```
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36
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+
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+
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+
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39
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+
別解釈として、CCCが有効(Nullでない)なAAAとBBBのパターンの件数を求めるという事だと、
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40
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```SQL
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42
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select AAA,CCC,(select count(*) from テーブル where AAA=V1.AAA and BBB=V1.BBB) cnt
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from (
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46
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select AAA,BBB,CCC
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48
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from テーブル t1
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50
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+
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51
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where CCC is not null
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+
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group by AAA,BBB,CCC
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54
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55
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) v1
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56
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+
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57
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```
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2
追記
test
CHANGED
@@ -17,3 +17,19 @@
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17
17
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group by AAA,CCC
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18
18
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19
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```
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+
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21
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> CCCはどれか一つだけ必ず値あり
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22
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+
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+
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25
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ということなので、以下のようにするだけでも良いかも
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```SQL
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+
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select AAA,CCC,(select count(*) from テーブル where BBB=t1.BBB) cnt
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30
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+
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31
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from テーブル t1
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32
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+
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33
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where CCC is not null
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34
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+
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35
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```
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1
訂正
test
CHANGED
@@ -1,12 +1,18 @@
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1
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-
相関副問合せで件数を取得します
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1
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+
相関副問合せで件数を取得しておき、それを集計します
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2
2
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3
3
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```SQL
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4
4
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5
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-
select AAA,CCC,(
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5
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+
select AAA,CCC, SUM(cnt)
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6
6
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7
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-
from
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7
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from (
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8
8
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9
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select AAA,CCC,(select count(*) from テーブル where BBB=t1.BBB) cnt
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10
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+
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11
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from テーブル t1
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12
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+
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9
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where CCC is not null
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13
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where CCC is not null
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14
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+
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15
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+
) iv
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10
16
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11
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group by AAA,CCC
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12
18
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