我有一个mysqli查询，我需要将其格式化为适用于移动应用程序的JSON。
我已经成功生成了一个查询结果的XML文档，但我正在寻找更轻量级的解决方案。（请参见下面的当前XML代码）
2
867$mysql = new mysqli(DB_SERVER,DB_USER,DB_PASSWORD,DB_NAME) or die('连接数据库时出现问题');
$stmt = $mysql->prepare('SELECT DISTINCT title FROM sections ORDER BY title ASC');
$stmt->execute();
$stmt->bind_result($title);
// 创建xml格式
$doc = new DomDocument('1.0');
// 创建根节点
$root = $doc->createElement('xml');
$root = $doc->appendChild($root);
// 为每一行添加节点
while($row = $stmt->fetch()) :
$occ = $doc->createElement('data');
$occ = $root->appendChild($occ);
$child = $doc->createElement('section');
$child = $occ->appendChild($child);
$value = $doc->createTextNode($title);
$value = $child->appendChild($value);
endwhile;
$xml_string = $doc->saveXML();
header('Content-Type: application/xml; charset=ISO-8859-1');
// 输出xml，jQuery准备就绪
echo $xml_string;
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这是我创建JSON feed的方法：
$mysqli = new mysqli('localhost', 'user', 'password', 'myDatabaseName'); $myArray = array(); if ($result = $mysqli->query("SELECT * FROM phase1")) { $tempArray = array(); while ($row = $result->fetch_object()) { $tempArray = $row; array_push($myArray, $tempArray); } echo json_encode($myArray); } $result->close(); $mysqli->close();只需从查询结果创建一个数组，然后对其进行编码
$mysqli = new mysqli('localhost','user','password','myDatabaseName'); $myArray = array(); $result = $mysqli->query("SELECT * FROM phase1"); while($row = $result->fetch_assoc()) { $myArray[] = $row; } echo json_encode($myArray);输出结果如下：
[ {"id":"31","name":"product_name1","price":"98"}, {"id":"30","name":"product_name2","price":"23"} ]如果你想要另一种样式，可以将fetch_assoc()改为fetch_row()，得到如下输出：
[ ["31","product_name1","98"], ["30","product_name2","23"] ]