質問編集履歴
3
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@@ -106,7 +106,7 @@
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// 結果を取得する
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$
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$aa = $this -> AA-> get($id, ['contain' => ['KbnMasters']]);
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}
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@@ -4,7 +4,7 @@
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4
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そのため、belongsToを使用し、発行したSQLが別名となるようにしたい
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現在発行されるSQLだと//
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現在発行されるSQLだと//km_Aの箇所が上書きされて//km_Bの箇所のSQLのみが発行されてしまう
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@@ -14,51 +14,33 @@
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###テーブル構成
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studios（店舗マスタ）
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-
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id|studio_nm|studio_kbn|apply_kbn
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-
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kbn_masters（区分マスタ）
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id|subjectid|subject_name|kbnid|kbn_name
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店舗マスタのstudio_kbnに対応する区分マスタはsubjectidが3のもので、apply_kbnに対応する区分マスタはsubjectidが4となります。
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SQL取得イメージ
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```sql
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select
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-
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AA.*
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, km_
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, km_A.kbn_name
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, km_
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, km_B.kbn_name
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from
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-
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AA
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left outer join kbn_master km_
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left outer join kbn_master km_A
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on km_
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on km_A.subject_id = 3
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and km_
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and km_A.kbn_id = AA.c_kbn
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left outer join kbn_master km_
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left outer join kbn_master km_B
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40
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on km_
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on km_B.subject_id = 4
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and km_
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and km_B.kbn_id = AA.d_kbn
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;
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@@ -82,25 +64,25 @@
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//
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//km_A
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$this -> belongsTo('KbnMasters', [
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'bindingKey' => 'kbnid',
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'foreignKey' => '
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'foreignKey' => 'km_A',
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'conditions'=>['subjectid' => '3' ],
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'joinType' => 'LEFT']);
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//
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//km_B
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$this -> belongsTo('KbnMasters', [
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'bindingKey' => 'kbnid',
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'foreignKey' => '
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'foreignKey' => 'km_B',
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'conditions'=>['subjectid' => '4' ],
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1
初心者マークを追加
test
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test
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