我想要html格式的这段代码
<a href="images/p191-large.jpg" class="cloud-zoom-gallery" rel="useZoom: 'zoom11', smallImage: 'images/p191.jpg' " title="Women's Crepe Printed Black">
<img class="zoom-tiny-image" itemprop="image" src="images/p191.jpg" width="80" height="97" alt=""/>
</a>
我已经用 cakephp 编写了这段代码以获得上面的结果
<?php
echo $this->Html->link(
$this->Html->image("p191.jpg", array(
"alt" => "Pant Suit",
"width" => "80",
"height" => "97",
'class' => 'zoom-tiny-image'
)),
array(
'controller' => 'admins',
'action' => '$this->Html->image(p191.jpg)'
),
array(
'class' => 'cloud-zoom-gallery',
'title' => 'Women''s Crepe Printed Black',
'escape' => false,
'rel' => "useZoom: 'zoom11', smallImage: 'images/p191.jpg' "
)
);
?>
但是将其作为输出
<a href="/stylishtailor1/admins/ ;image(p191.jpg)" class="cloud-zoom-gallery" title="Women's Crepe Printed Black" rel="useZoom: 'zoom11', smallImage: 'images/p191.jpg' ">
<img src="/stylishtailor1/images/p191.jpg" alt="Pant Suit" width="80" height="97" class="zoom-tiny-image" />
</a>
我该怎么做，请回复???
这是最接近您想要的代码：
<?php
echo $this->Html->link(
$this->Html->image('/images/p191-large.jpg', array('alt' => 'Pant Suit', 'width' => '80', 'height' => '97', 'class' => 'zoom-tiny-image')),
'images/p191-large.jpg',
array('class' => 'cloud-zoom-gallery', 'title' => 'Women''s Crepe Printed Black', 'escape' => false, 'rel' => "useZoom: 'zoom11', smallImage: 'images/p191.jpg' ")
);
?>
为了确保它正常工作，您必须检查两件事。图像路径（image()函数的第一个参数）应该具有完整路径，如果它不在webroot/img下，或者如果它在那里，可以省略起始/。
要检查的第二件事是您的 url 位置（link()函数中的第二个参数）。您可以将其作为字符串或路由数组传递。在你的例子中，你传递了非常奇怪的数组。