Question

我的数据库表：

TABLE `events` (
    `event_id` INT(11) unsigned NOT NULL AUTO_INCREMENT,
    `event_title` VARCHAR(255) NOT NULL,
    `event_desc` TEXT,
    `event_location` VARCHAR(255) NOT NULL,
    `event_requirements` TEXT DEFAULT NULL,
    `event_date` DATETIME NOT NULL,
 PRIMARY KEY (`event_id`)

目标：在选择下拉菜单中选择事件时，我想显示其余的数据库数据，这些数据属于下面div中的事件。

尝试解决问题（在MVC架构中）：

控制器

  function index()
  {
      $overview_model = $this->loadModel('Events');
      $this->view->events = $overview_model->getEventTypes();
      $this->view->render('events/index');
  }

模型

  public function getEventTypes()
  {
      $sth = $this->db->prepare("SELECT * FROM events");
      $sth->execute();

      $events = array();

      foreach ($sth->fetchAll() as $event) {
          $events[$event->event_id] = new stdClass();
          $events[$event->event_id]->event_id = $event->event_id;
          $events[$event->event_id]->event_title = $event->event_title;
          $events[$event->event_id]->event_desc = $event->event_desc;
          $events[$event->event_id]->event_location = $event->event_location;
          $events[$event->event_id]->event_requirements = $event->event_requirements;
          $events[$event->event_id]->event_date = $event->event_date;
      }

      return $events;
  }

查看

    <!-- start event type selection --> 
   <div class="event-selection">
     <label>Choose an event:</label>
     <select id="select-event-type">
        <?php foreach ($this->events as $event) {
            // Using the event's ID as a value makes it easier for us to look up the selected event.
            echo "<option value='" .$event->event_id. "'>" .$event->event_title."</option>";
        }?>
    </select>

    <script>
        (function ($) {
            // Here we set the JavaScript variable 'events' to a JSON-encoded copy of the array we passed to the template, or null (if there are no events).
            var events = <?php echo (count($this->events) > 0) ? "JSON.parse(" .json_encode($this->events). ");" : "null"; ?>
            $(document).ready(function() {
                $('#select-event-type').change(function() {
                    if (events) {
                        var event = events[this.value];
                        $('#event-details').html(event.title);
                    }
                });
            });
        })($);
    </script>   

    <!-- Start event-details -->
    <div id="event-details"></div> <!-- /#event-details -->

</div><!-- /.event-selection -->

在页面上，当我更改下拉选项时没有任何反应。

控制台输出（在chrome中）：

        <script>
            (function ($) {
                var events = JSON.parse({"1":{"event_id":"1","event_title":"event1","event_desc":"event1","event_location":"eventlocation","event_requirements":"event1","event_date":"2022-07-20 15:00:00"},"2":{"event_id":"2","event_title":"event2","event_desc":"event2","event_location":"eventlocation","event_requirements":"event2","event_date":"2015-04-20 15:00:00"},"3":{"event_id":"3","event_title":"event3","event_desc":"event3","event_location":"eventlocation","event_requirements":"event3","event_date":"2019-11-20 16:00:00"}});             $(document).ready(function() {
                    $('#select-event-type').change(function() {
                        if (events) {
                            // Now we can get the event details from the selected value.
                            var event = events[this.value];
                            $('#event-details').html(event.title);
                        }
                    });
                });
            })($);
        </script>

错误：

未捕获的SyntaxError：意外的令牌o

根据我的研究，产生错误的一行就是这个（在视图中）：

var events = <?php echo (count($this->events) > 0) ? "JSON.parse(" .json_encode($this->events). ");" : "null"; ?>

控制台中的JSON格式是否正确？另一个线程建议使用$ .getJSON（）而不是JSON.parse（）但是，因为这是我第一次使用JSON，所以我不太确定在我的情况下这是如何工作的。
我非常感谢任何帮助！

Answer 1

var events = JSON.parse({"1":{"…

您正在尝试解析JavaScript对象文字，就好像它是JSON一样。

由于你必须有一个字符串才能解析JSON，它会被转换成一个字符串给你相当于：

var events = JSON.parse("[object Object]");

那不是JSON。

直接使用对象文字：

var events = {"1":{"…

与JSON相关的错误：意外的令牌o

1 个答案: