使用Cake \ ORM \ TableRegistry;
$ websites => TableRegistry ::得到( '网站');
$ query => $ websites-> find() - > order(['creation_date'=>'DESC']);
echo "<table>"; echo"<tr> <td>Cover Image</td> <td>Company Name</td> <td>Date Added</td> </tr>"; foreach ($query as $row) { echo"<tr> <td>".$row['Website']['image']."</td> <td>".$row->company."</td <td>".$row->creation_date."</td> </tr>"; } echo"</table>"; ?>
I remember this code in cakephp2 and display the images path "app/webroot/img/websites/"
<?php echo $this->Html->image('websites/' . $row['Website']['image']);
但如何在cakephp3中执行此操作,以显示图像。 ?&GT;
答案 0 :(得分:0)
在cakephp 3中没有任何大的区别,它很简单,如下面的代码
<?php foreach ($websites as $websites): ?>
<?= $this->Html->image('websites/'.$websites->image) ?>
<?php endforeach; ?>
答案 1 :(得分:0)
您似乎不知道自己的真正问题。您的问题似乎是有人在用正确的HTML代码上传图片时遇到问题。为此,您只需要在数据库表列中为varchar类型指定该上载字段,然后在视图中将图像从文件类型上载至该字段即可。像这样:
<div class="form-group">
<?php
echo $this->Form->input('your_column_name', [ 'type' => 'file']);
?>
</Div>
但是,如果您的问题是有关无法检索或发送到index.ctp或前端视图的上载图像,那么您的解决方案就在这里: How to retrieve in cakephp3.7 the name as string of an image uploaded trough a form with blob column into a mysql database?