回答編集履歴
2
修正
answer
CHANGED
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@@ -11,13 +11,9 @@
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11
11
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select t1.up_time, t1.picupno, t2.sum_qty
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12
12
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from test as t1
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13
13
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inner join (
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14
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-
select sub_id, sum(qty) as sum_qty from test group by sub_id
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14
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+
select sub_id, max(up_time) as max_up_time, sum(qty) as sum_qty from test group by sub_id
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15
15
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) as t2
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16
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-
on t1.sub_id=t2.sub_id
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16
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+
on t1.sub_id=t2.sub_id and t1.up_time=t2.max_up_time
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17
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-
inner join (
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18
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-
select sub_id, max(up_time) as sum_qty from test group by sub_id
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19
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-
) as t2
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20
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-
on t1.sub_id=t3.sub_id
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21
17
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```
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22
18
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MySQLはサンプルデータを触っている程度なので、大量のデータでのSQLの記述による性能差などは見地がありません。
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23
19
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なので、結果をフィードバックして頂けると喜びます。
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1
追記
answer
CHANGED
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@@ -5,4 +5,19 @@
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5
5
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,(select sum(qty) from test where sub_id=t1.sub_id) as sum_qty
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6
6
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from test as t1
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7
7
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where up_time=(select max(up_time) from test where sub_id=t1.sub_id)
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8
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-
```
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8
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+
```
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9
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+
一応別解（相関問い合わせによるサブクエリーを使用しない版）
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10
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+
```SQL
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11
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+
select t1.up_time, t1.picupno, t2.sum_qty
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|
12
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+
from test as t1
|
|
13
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+
inner join (
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|
14
|
+
select sub_id, sum(qty) as sum_qty from test group by sub_id
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15
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+
) as t2
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16
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+
on t1.sub_id=t2.sub_id
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17
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+
inner join (
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18
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+
select sub_id, max(up_time) as sum_qty from test group by sub_id
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19
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+
) as t2
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20
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+
on t1.sub_id=t3.sub_id
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21
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+
```
|
|
22
|
+
MySQLはサンプルデータを触っている程度なので、大量のデータでのSQLの記述による性能差などは見地がありません。
|
|
23
|
+
なので、結果をフィードバックして頂けると喜びます。
|