如何使用php mysql获取嵌套的json对象

$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "database";

$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}

$sql = "SELECT data FROM users";
$result = $conn->query($sql);

if ($result->num_rows > 0) {
    while ($row = $result->fetch_assoc()) {
        $jsonData = json_decode($row['data'], true);
        // 处理嵌套的JSON对象
        // ...
    }
} else {
    echo "No results found";
}

{
    "name": "John",
    "age": 30,
    "address": {
        "street": "123 Main St",
        "city": "New York"
    }
}

$name = $jsonData['name'];
$age = $jsonData['age'];
$street = $jsonData['address']['street'];
$city = $jsonData['address']['city'];

echo "Name: " . $name . "<br>";
echo "Age: " . $age . "<br>";
echo "Address: " . $street . ", " . $city;

$conn->close();