PHP MySQL选择并作为json回显

Question

我如何回显mysql select as json的结果？当前它将回显响应：

ID: 1 - Name: John Doe

ID: 2 - Name: John Deo

此致，

$conn = new mysqli($servername, $username, $password, $dbname);

// Check connection

if ($conn->connect_error) {

    die("Connection failed: " . $conn->connect_error);

} 

$sql = "SELECT id, firstname, lastname FROM MyGuests";

$result = $conn->query($sql);

if ($result->num_rows > 0) {

    // output data of each row

    while($row = $result->fetch_assoc()) {

        echo "id: " . $row["id"]. " - Name: " . $row["firstname"]. " " . $row["lastname"]. "<br>";

    }

} else {

    echo "0 results";

}

$conn->close();

Answer 1

首先将所有行添加到一个数组中：$data = [];while($row = $result->fetch_assoc()) {&nbsp; &nbsp; $data[] = $row;}然后将数组转换为json并输出：echo json_encode($data);