事象

• 歯抜け添字の配列はJSON変換後に連想配列（JSONではオブジェクト）になる
• 連番添字の配列はJSON変換後に配列になる

添字が歯抜けている場合

$items = [
  0 => ['key' => 'name'],
  1 => ['key' => 'post_code'],
  2 => ['key' => 'address'],
  3 => ['key' => 'contact'],
  4 => ['key' => 'age'],
  // 5〜7が抜けている（歯抜け添字）
  8 => ['key' => 'previous_place_of_stay'],
  9 => ['key' => 'next_place_of_stay'],
];

echo json_encode($items, JSON_UNESCAPED_UNICODE) . PHP_EOL;

出力

{
  "0": {"key": "name"},
  "1": {"key": "post_code"},
  "2": {"key": "address"},
  "3": {"key": "contact"},
  "4": {"key": "age"},
  "8": {"key": "previous_place_of_stay"},
  "9": {"key": "next_place_of_stay"}
}

添字が連番の場合

$items = [
  0 => ['key' => 'name'],
  1 => ['key' => 'post_code'],
  2 => ['key' => 'address'],
  3 => ['key' => 'contact'],
  4 => ['key' => 'age'],
  5 => ['key' => 'previous_place_of_stay'],
  6 => ['key' => 'next_place_of_stay'],
];

echo json_encode($items, JSON_UNESCAPED_UNICODE) . PHP_EOL;

出力

[
  {"key": "name"},
  {"key": "post_code"},
  {"key": "address"},
  {"key": "contact"},
  {"key": "age"},
  {"key": "previous_place_of_stay"},
  {"key": "next_place_of_stay"}
]

原因

• json_encodeは添字が連番ではない配列を「オブジェクト」として扱う
• 歯抜けキーの配列はJSON化時にオブジェクトとして出力される

解決方法

array_valuesで添字を詰めてから返す。

$items = array_values($items);

echo json_encode($items, JSON_UNESCAPED_UNICODE) . PHP_EOL;

出力

[
  {"key": "name"},
  {"key": "post_code"},
  {"key": "address"},
  {"key": "contact"},
  {"key": "age"},
  {"key": "previous_place_of_stay"},
  {"key": "next_place_of_stay"}
]

まとめ

• 歯抜け添字の配列はJSON変換後に連想配列（オブジェクト）になる
• 連番添字の配列はJSON変換後に配列になる
• 歯抜け添字の配列を配列として返すにはarray_valuesで整形する