我正在尝试解决如何从服务器端文件返回base64结果,并将其显示在现有的img标记中。
我有下面的php是显示第一个img ok -但第二个显示为破碎。
这是怎么回事
<?php
$smiley="data:image/png;base64,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";
if ( isset($_GET['img']) && $_GET['img'] === 'Y' ) {
echo $smiley;
exit;
}
echo "<img src='$smiley'> ";
$http=$_SERVER['REQUEST_SCHEME']."://";
$siteurl=$_SERVER['HTTP_HOST'];
$sitepage=$_SERVER['PHP_SELF'];
echo "<img src='$http$siteurl$sitepage?img=Y'>";
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2条答案
按热度按时间lawou6xi1#
好吧,我解决了:
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yuvru6vn2#
PHP支持数据URI:
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当上下文是超文本中图像的src时,您不需要发送content-type。