回答編集履歴
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sample
answer
CHANGED
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@@ -1,4 +1,58 @@
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1
1
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おそらくCONCATはしないほうがいいと思います
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2
2
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RDBからの戻りはassocでfetchAllすればスプレッドシート的な
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3
3
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縦×横のデータになるので、組み替えてからjsonエンコードするだけです
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4
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-
組み換えについては元データと結果の提示がないのでなんとも言えません
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4
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組み換えについては元データと結果の提示がないのでなんとも言えません
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5
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6
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# sample
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```PHP
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9
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/*$aはfetchAllするものとします*/
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10
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$a=[
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11
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[
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12
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"type"=>"t_1",
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13
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"properties01"=>"p1_1",
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14
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"properties02"=>"p2_1",
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15
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"material01_shop01"=>"m1s1_1",
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"material01_shop02"=>"m1s2_1",
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"material02_shop01"=>"m2s1_1",
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"material02_shop02"=>"m2s2_1",
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"id"=>"i_1",
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],
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[
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"type"=>"t_2",
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"properties01"=>"p1_2",
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"properties02"=>"p2_2",
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"material01_shop01"=>"m1s1_2",
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"material01_shop02"=>"m1s2_2",
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"material02_shop01"=>"m2s1_2",
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"material02_shop02"=>"m2s2_2",
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"id"=>"i_2",
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],
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[
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"type"=>"t_3",
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"properties01"=>"p1_3",
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"properties02"=>"p2_3",
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"material01_shop01"=>"m1s1_3",
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"material01_shop02"=>"m1s2_3",
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"material02_shop01"=>"m2s1_3",
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"material02_shop02"=>"m2s2_3",
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"id"=>"i_3",
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40
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],
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];
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$b=[];
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foreach($a as $key=>$val){
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$b[$key]["type"]=$val["type"];
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$b[$key]["properties"]=[];/*空配列は宣言しなくてもOKだが念の為*/
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$b[$key]["properties"]["properties01"]=$val["properties01"];
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47
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$b[$key]["properties"]["properties02"]=$val["properties02"];
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48
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$b[$key]["material"]=[];
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49
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$b[$key]["material"]["material01"]=[];
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50
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$b[$key]["material"]["material01"]["material01_shop01"]=$val["material01_shop01"];
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51
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$b[$key]["material"]["material01"]["material01_shop02"]=$val["material01_shop02"];
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$b[$key]["material"]["material02"]=[];
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$b[$key]["material"]["material02"]["material02_shop01"]=$val["material02_shop01"];
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$b[$key]["material"]["material02"]["material02_shop02"]=$val["material02_shop02"];
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$b[$key]["idl"]=$val["id"];
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}
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print_r(json_encode($b, JSON_PRETTY_PRINT));
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58
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```
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