php - 如何将 mysqli 结果转换为 JSON？

Question

我有一个 mysqli 查询，我需要为移动应用程序格式化为 JSON。

我已经成功地为查询结果生成了一个 XML 文档，但是我正在寻找更轻量级的东西。 (下面是我当前的 XML 代码)

$mysql = new mysqli(DB_SERVER,DB_USER,DB_PASSWORD,DB_NAME) or die('There was a problem connecting to the database');

$stmt = $mysql->prepare('SELECT DISTINCT title FROM sections ORDER BY title ASC');
$stmt->execute();
$stmt->bind_result($title);

// create xml format
$doc = new DomDocument('1.0');

// create root node
$root = $doc->createElement('xml');
$root = $doc->appendChild($root);

// add node for each row
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');

// output xml jQuery ready

echo $xml_string;

Answer 1

只需根据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);

输出如下:

[
    {"id":"31","name":"product_name1","price":"98"},
    {"id":"30","name":"product_name2","price":"23"}
]

如果你想要其他样式，可以将 fetch_assoc() 更改为 fetch_row() 并获得如下输出:

[
    ["31","product_name1","98"],
    ["30","product_name2","23"]
]