我如何从图像文件夹(权限644)在php服务器上读取文件，并将其动态设置为html的img src ?我尝试了以下
<?php
error_reporting(E_ALL);
ini_set('display_errors', 1);
$name = './images/approved.jpg';
//$fp = readfile($name);
$im = imagecreatefromjpeg($name);
echo "<img src='data:image/jpg;base64,+ 'base64_encode($im), alt='Approved Videos''>";
exit;
?>
我也试过下面的
<div id="videoThumbnails">
<p>Thumbnail</p>
<a href="#" class="thumbnail">
<!-- <img src="/phpVA/thumbnails/a.mp4.jpg" alt="Denied Videos">-->
<?php
$name="/phpVA/thumbnails/a.mp4.jpg";
$im = file_get_contents($name);
echo "<img src='data:image/jpg;base64,".base64_encode($im)."' alt='Approved Videos'>";
?>
</a>
</div>
输出
<img src="data:image/jpg;base64," alt="Approved Videos">
尝试file_get_contents
$im = file_get_contents($name);
echo "<img src='data:image/jpg;base64,".base64_encode($im)."' alt='Approved Videos'>";
我建议使用file_get_contents来获取文件内容，并用base64对这些内容进行编码。下面是我在我的一个项目中使用的一个很好的例子:
$path = './images/approved.jpg';
$type = pathinfo($path, PATHINFO_EXTENSION);
$data = file_get_contents($path);
$base64 = 'data:image/' . $type . ';base64,' . base64_encode($data);