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 DoeID: 2 - Name: John Deo此致，$conn = new mysqli($servername, $username, $password, $dbname);// Check connectionif ($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()) {

    $data[] = $row;

}

然后将数组转换为json并输出：

echo json_encode($data);