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）
$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; 
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：