Question

我目前正在尝试将YouTube中的信息插入数据库以供将来使用。当插入进入并且闪存返回时，我没有抛出任何错误，就像数据插入成功一样。问题是只有一列数据插入到表中。这是我正在使用的代码。

观点：

<?php foreach ($videos->data->items as $video) : ?>
        <?php echo $this->Form->create('Youtube', array('action' => 'addNews')); ?>
        <tr>
            <td><?php echo $this->Html->image($video->thumbnail->hqDefault, array('width' => 72, 'height' => 56, 'alt' => 'image')); ?></td>
            <td><?php echo $this->Time->niceShort($video->uploaded); ?></td>
            <td><?php echo $video->category; ?></td>
            <td><?php echo $video->title; ?></td>
            <td><?php echo $this->Text->truncate($video->description, 50); ?></td>
            <td><?php echo $this->Number->precision($video->rating, 2); ?></td>
            <?php echo $this->Form->input('youtube_url', array('type' => 'hidden', 'value' => $video->id)); ?>
            <?php echo $this->Form->input('uploaded', array('type' => 'hidden', 'value' => $this->Time->format('Y-m-d H:i:s', $video->uploaded))); ?>
            <?php echo $this->Form->input('updated', array('type' => 'hidden', 'value' => $this->Time->format('Y-m-d H:i:s', $video->updated))); ?>
            <?php echo $this->Form->input('uploader', array('type' => 'hidden', 'value' => $video->uploader)); ?>
            <?php echo $this->Form->input('category', array('type' => 'hidden', 'value' => $video->category)); ?>
            <?php echo $this->Form->input('title', array('type' => 'hidden', 'value' => $video->title)); ?>
            <?php echo $this->Form->input('description', array('type' => 'hidden', 'value' => $video->description)); ?>
            <?php echo $this->Form->input('sqDefault', array('type' => 'hidden', 'value' => $video->thumbnail->sqDefault)); ?>
            <?php echo $this->Form->input('hqDefault', array('type' => 'hidden', 'value' => $video->thumbnail->hqDefault)); ?>
            <?php echo $this->Form->input('default_player', array('type' => 'hidden', 'value' => $video->player->default)); ?>
            <?php echo $this->Form->input('mobile_player', array('type' => 'hidden', 'value' => $video->player->mobile)); ?>
            <?php echo $this->Form->input('duration', array('type' => 'hidden', 'value' => $video->duration)); ?>
            <?php echo $this->Form->input('rating', array('type' => 'hidden', 'value' => $video->rating)); ?>
            <?php echo $this->Form->input('like_count', array('type' => 'hidden', 'value' => $video->likeCount)); ?>
            <?php echo $this->Form->input('view_count', array('type' => 'hidden', 'value' => $video->viewCount)); ?>
            <?php echo $this->Form->input('comment_count', array('type' => 'hidden', 'value' => $video->commentCount)); ?>
            <?php echo $this->Form->input('embed', array('type' => 'hidden', 'value' => $video->accessControl->embed)); ?>
            <td><?php echo $video->viewCount; ?></td>
            <td><?php echo $video->accessControl->embed; ?></td>
            <td><?php echo $this->Form->submit('Add News'); ?></td>
        </tr>
        <?php //echo $this->Form->end(); ?>
    <?php endforeach; ?>

我认为可能有一种更好的方式来显示项目并设置值，虽然我是新蛋糕所以我不确定我应该寻找什么，以便制作一个预设不变的值其他形式的表单而不是form-＆gt; input（type =＆gt; hidden）;

控制器：

if($this->request->is('post') || $this->request->is('put')) {

            //debug($this->request->data);
            if($this->Youtube->save($this->request-data)) {
                $this->Session->setFlash('Your post was added');
                $this->redirect($this->referer());
            }
        }

我已经检查了我的表等命名，但它们都是正确的。我不知道中断发生在哪里，或者我怎么能得到蛋糕来传递正在发生的任何MySQL错误。任何帮助将不胜感激。

Answer 1

愚蠢在这一方面得到了最好的...语法拼写错误导致我无法在此处保存数据：if($this->Youtube->save($this->request-data))应该是if($this->Youtube->save($this->request->data))。错过了一个> ......我想有一半的时间就是问题，但这是一个错误的错误！

数据未插入CakePHP

1 个答案: