回答編集履歴
3
修正
answer
CHANGED
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@@ -1,6 +1,6 @@
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1
1
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条件に一致する場合を1として合計すれば良いかと思います。
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2
2
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```SQL
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3
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-
select A.CODE, sum(iif(A.Afield < B.Bfield, 1, 0))
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3
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+
select A.CODE, sum(iif(A.Afield < B.Bfield, 1, 0)) as cnt
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4
4
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from tableA as A inner join tableB as B
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5
5
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on A.CODE=B.CODE
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6
6
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group by A.CODE
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@@ -9,7 +9,7 @@
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9
9
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10
10
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上記はすべてのデータの一覧なので、コードをピンポイントで指定した場合は、group byは不要です。
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11
11
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```SQL
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12
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-
select sum(iif(A.Afield < B.Bfield, 1, 0))
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12
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+
select sum(iif(A.Afield < B.Bfield, 1, 0)) as cnt
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13
13
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from tableA as A inner join tableB as B
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14
14
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on A.CODE=B.CODE
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15
15
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where A.CODE=111
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2
修正
answer
CHANGED
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@@ -3,6 +3,14 @@
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3
3
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select A.CODE, sum(iif(A.Afield < B.Bfield, 1, 0))
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4
4
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from tableA as A inner join tableB as B
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5
5
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on A.CODE=B.CODE
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6
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+
group by A.CODE
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7
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+
```
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8
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+
※```on A.CODE=B.CODE```は結合条件なので、実際とは違うかもしれません
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9
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+
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10
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+
上記はすべてのデータの一覧なので、コードをピンポイントで指定した場合は、group byは不要です。
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11
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+
```SQL
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12
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+
select sum(iif(A.Afield < B.Bfield, 1, 0))
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13
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+
from tableA as A inner join tableB as B
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14
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+
on A.CODE=B.CODE
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6
15
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where A.CODE=111
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7
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-
group by A.CODE
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8
16
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```
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1
修正
answer
CHANGED
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@@ -3,5 +3,6 @@
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3
3
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select A.CODE, sum(iif(A.Afield < B.Bfield, 1, 0))
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4
4
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from tableA as A inner join tableB as B
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5
5
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on A.CODE=B.CODE
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6
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+
where A.CODE=111
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6
7
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group by A.CODE
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7
8
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```
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