質問編集履歴
3
php manualに載ってる方法を試してみましたがエラーが出ました
title
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File without changes
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body
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@@ -6,4 +6,14 @@
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6
6
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//実行結果
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7
7
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array(1) { [0]=> array(1) {
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8
8
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["array_json"]=> string(42) "{"bg-blue1":"a","bg-blue2":"b","bg-blue3":"c"}"
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9
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-
} }
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9
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+
} }
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10
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+
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11
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+
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12
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+
試したこと
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13
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$results_decode = json_decode($results);
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14
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print $results_decode->{'bg-blue3'};
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15
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+
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//実行結果
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17
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Warning: json_decode() expects parameter 1 to be string, array given in line 26
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18
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NULL
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19
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Notice: Trying to get property 'bg-blue3' of non-object in line 29
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2
タイトルを変えました
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@@ -1,1 +1,1 @@
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1
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-
[PHP]
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1
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+
[PHP]取得したJSONファイルの中身を取り出したい
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body
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File without changes
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1
余計にエンコードしてたのを治しました
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File without changes
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body
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@@ -2,9 +2,7 @@
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2
2
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$stmt = $db->prepare("SELECT array_json FROM draft_data WHERE id=1");
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3
3
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$stmt->execute();
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4
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$results = $stmt->fetchAll(PDO::FETCH_ASSOC);
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5
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-
$json = json_encode($results);
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6
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-
$value = json_decode($json,true);
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7
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-
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5
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+
var_dump($results);
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8
6
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//実行結果
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9
7
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array(1) { [0]=> array(1) {
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10
8
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["array_json"]=> string(42) "{"bg-blue1":"a","bg-blue2":"b","bg-blue3":"c"}"
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