如何将mysqli查询结果转换为JSON格式？

Question

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;

Answer 1

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();

Answer 2

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"]
]