如何將mysqli查詢結果轉換為JSON格式？

Question

我有一個mysqli查詢，我需要將其格式化為適用於行動應用程式的JSON。

我已經成功產生了一個查詢結果的XML文檔，但我正在尋找更輕量級的解決方案。 （請參閱下面的當前XML程式碼）

$mysql = new mysqli(DB_SERVER,DB_USER,DB_PASSWORD,DB_NAME) 或 die('連接資料庫時出現問題');

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

// 建立xml格式
$doc = new DomDocument('1.0');

// 建立根節點
$root = $doc->createElement('xml');
$root = $doc->appendChild($root);

// 為每一行新增節點
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');

// 輸出xml，jQuery準備就緒

echo $xml_string;

Answer 1

這是我創建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

只需從查詢結果建立一個數組，然後對其進行編碼

$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);

輸出結果如下：

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

如果你想要另一種樣式，可以將fetch_assoc()改為fetch_row()，得到以下輸出：

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