回答編集履歴
2
追記
answer
CHANGED
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@@ -16,4 +16,40 @@
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16
16
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facilities_trait=(flag, type, facilities_id, rooms_id)
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rooms=(flag, status, articles_id, id)
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articles=(flag, id)
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19
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-
※項目の並びは、カーディナリティに合わせて適宜変更が必要です。
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19
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+
※項目の並びは、カーディナリティに合わせて適宜変更が必要です。
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20
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+
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21
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追記
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22
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--
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23
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※SQLが変更されたので、内容に合わせて追記
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24
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素直に記述。
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25
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```SQL
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26
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SELECT * FROM facilities_trait
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27
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WHERE flag = 2 AND type in (1, 3)
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28
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AND trait_id in (
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29
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SELECT id FROM articles
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30
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WHERE flag = 2
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31
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AND id in (
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32
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SELECT articles_id FROM rooms
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33
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WHERE flag = 2 AND status = 2
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34
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AND id in (
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35
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SELECT trait_id FROM facilities_trait
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36
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WHERE facilities_id = 31
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37
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)
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38
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)
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39
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)
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40
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```
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41
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条件内をJoinで。
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42
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```SQL
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43
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SELECT * FROM facilities_trait
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44
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WHERE flag = 2 AND type in (1, 3)
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45
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AND trait_id in (
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46
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select articles.id
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47
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FROM articles
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48
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inner join rooms
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49
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on articles.id = rooms.articles_id
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50
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and articles.flag = 2 and room.flag = 2 AND room.status = 2
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51
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inner join facilities_trait
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52
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on rooms.id=facilities_trait.trait_id
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53
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and facilities_trait.facilities_id = 31
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54
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)
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55
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```
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1
推敲
answer
CHANGED
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@@ -14,6 +14,6 @@
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14
14
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```
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15
15
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それからそれぞれのテーブルに以下のインデックスがあれば、効率が良くなると思います。
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16
16
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facilities_trait=(flag, type, facilities_id, rooms_id)
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17
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-
rooms=(flag, status, id)
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17
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+
rooms=(flag, status, articles_id, id)
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18
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articles=(flag, id)
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19
19
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※項目の並びは、カーディナリティに合わせて適宜変更が必要です。
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