PHP json形式でデータを渡すテンプレ

PHP json形式でデータを渡すテンプレ
header("Content-Type: application/json; charset=UTF-8");

// 記事一覧を得る
$Get = 'SELECT * FROM `sql_table` WHERE `PageShow` = 1 ORDER BY `UpDateTime` DESC';
$Result = mysqli_query($MyLink, $Get);
//配列を用意
$ResultArr = [];
// 連続表示処理
while($Data = mysqli_fetch_object($Result)){
array_push(
	$ResultArr,[
		"CateID" => $Data->CateID,
		"PageShow" => $Data->PageShow,
		"Holiday" => $Data->Holiday,
		"StartDate" => $Data->StartDate,
		"Title" => $Data->Title,
		"SubTitle" => $Data->SubTitle,
		"StartDateTime" => $Data->StartDateTime,
		"EndDateTime" => $Data->EndDateTime,
		"Item01" => $Data->Item01,
		"Item02" => $Data->Item02,
		"Item03" => $Data->Item03,
		"Item04" => $Data->Item04,
		"Item05" => $Data->Item05
	]);
}
mysqli_close($MyLink);
echo json_encode($ResultArr);

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