回答編集履歴
3
SQL修正
answer
CHANGED
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@@ -13,8 +13,7 @@
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13
13
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```SQL
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14
14
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SELECT table1.table1_id, table1.table1_name
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15
15
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FROM table1 INNER JOIN (
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16
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-
SELECT table1_id FROM table12 WHERE table2_id IN
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16
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+
SELECT table1_id FROM table12 WHERE table2_id IN (1,2)
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17
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-
(SELECT table2_id FROM table2 WHERE table2_id in (1, 2))
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18
17
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GROUP BY table1_id
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19
18
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HAVING count(*)=2
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20
19
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) vt1 ON (table1.table1_id = vt1.table1_id)
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2
誤字修正・table2 の WHERE 条件を id 由来に変更
answer
CHANGED
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@@ -5,7 +5,7 @@
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5
5
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SELECT DISTINCT table1.table1_id, table1.table1_name
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6
6
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FROM table1 INNER JOIN table12 ON (table12.table1_id = table1.id)
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7
7
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INNER JOIN table2 ON (table2.id = table12.table2_id)
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8
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-
WHERE table2.
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8
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+
WHERE table2.table2_id in (1, 2);
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9
9
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```
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10
10
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でいいですね。
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11
11
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@@ -14,7 +14,7 @@
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14
14
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SELECT table1.table1_id, table1.table1_name
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15
15
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FROM table1 INNER JOIN (
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16
16
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SELECT table1_id FROM table12 WHERE table2_id IN
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17
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-
(SELECT table2_id FROM table2 WHERE
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17
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+
(SELECT table2_id FROM table2 WHERE table2_id in (1, 2))
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18
18
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GROUP BY table1_id
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19
19
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HAVING count(*)=2
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20
20
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) vt1 ON (table1.table1_id = vt1.table1_id)
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1
誤字修正
answer
CHANGED
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@@ -2,7 +2,7 @@
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2
2
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3
3
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質問1 は
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4
4
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```SQL
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5
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-
SELECT DISTINCT table1.
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5
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+
SELECT DISTINCT table1.table1_id, table1.table1_name
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6
6
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FROM table1 INNER JOIN table12 ON (table12.table1_id = table1.id)
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7
7
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INNER JOIN table2 ON (table2.id = table12.table2_id)
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8
8
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WHERE table2.name in ('タマゴ', 'ケチャップ');
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