我错过了一些正确拉取图像的东西,但我无法弄清楚缺少什么。
echo "
<div class='"large-4'">
<img src='"images'"{$row['movies_fimg']}'" alt='"{$row['movies_title']}'">
<h2>{$row['movies_title']}</h2>
<p>{$row['movies_year']}</p>
<a href='"details.php?movie={$row['movies_id']}'">more...</a>
</div>
";
img 的 attr 中不需要双引号 src
<img src='"images'"{$row['movies_fimg']}'"
^^^
我认为这应该适合您:
(在 src 属性中添加了一个/
,所以如果你不需要它,请删除它)
echo "
<div class='large-4'>
<img src='images/" . $row['movies_fimg'] . "' alt='" . $row['movies_title'] . "'>
<h2>" . $row['movies_title'] . "</h2>
<p>" . $row['movies_year'] . "</p>
<a href='details.php?movie=" . $row['movies_id'] . "'>more...</a>
</div>
";
这里的 exta 引用:<img src='"images'"{$row['movies_fimg']}'"
正斜杠应该在哪里。
应该是:
echo "
<div class='"large-4'">
<img src='"images/{$row['movies_fimg']}'" alt='"{$row['movies_title']}'">
<h2>{$row['movies_title']}</h2>
<p>{$row['movies_year']}</p>
<a href='"details.php?movie={$row['movies_id']}'">more...</a>
</div>
";