我正在將一個json_encode對象返回給 ajax 調用。首先，有沒有更好的方法來做到這一點？這是json_encode needed?到我問題的根源。當我嘗試將鍵設為變量時，它會引發“未定義錯誤”。這一行：var displayTriggers = trigger_rows;有誰看到我做錯了什么？PHP：try {    $con = getConfig('pdo');    $con->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);    $sql_triggers = "        SELECT *        FROM triggers    ";    $triggers_stmt = $con->prepare($sql_triggers);    $triggers_stmt->execute();    $triggers_rows = $triggers_stmt->fetchAll(PDO::FETCH_ASSOC);    $triggers_arr = array();    foreach ($triggers_rows as $triggers_row) {        $trigger_id = $triggers_row['id'];        $trigger_title = $triggers_row['trigger_name'];        $trigger_setting = $triggers_row['setting'];        $trigger_user = $triggers_row['user_id'];        $trigger_placement = $triggers_row['placement'];        $trigger_date = $triggers_row['date_changed'];        $trigger_active = ( $trigger_setting == '1' ) ? ' active' : '';        $html = '';        $html .= '<div class="triggerRow" data-placement="'.$trigger_placement.'">';        $html .= '<div class="triggerRowLeft">';        $html .= '<div class="triggerTitle">' . $trigger_title . '</div>';        $html .= '<div class="triggerText">' . $trigger_date . '</div>';        $html .= '<div class="triggerText">' . $trigger_user . '</div>';        $html .= '</div>';        $html .= '<div class="triggerRowRight">';        $html .= '<div class="triggerButton' . $trigger_active . '"></div>';        $html .= '</div>';        $html .= '</div>';        $data = array('html' => $html);        $triggers_arr[] = $data;    }    echo json_encode(['trigger_rows' => $triggers_arr]);}catch(PDOException $e) {    echo "Connection failed: " . $e->getMessage();}