由于iOS6现在支持照片/视频上传,我想在我的手机应用程序中使用它。如果照片或视频都未上载到上载脚本提供的指定文件夹,则会出现问题。不确定我做错了什么。
下面是我用来抓取文件或捕获它的方法。
<form action="" method="POST">
<input type="file" accept="image/* capture="camera"><br><br>
<input type="submit" value="Upload" class="login-button"/>
</form>这将处理同一页上的照片...
<?php
function findexts ($filename)
{
$filename = strtolower($filename) ;
$exts = split("[/\\.]", $filename) ;
$n = count($exts)-1;
$exts = $exts[$n];
return $exts;
}
$ext = findexts ($_FILES['uploaded']['name']) ;
$ran = rand () ;
$ran2 = $ran.".";
$target = "files/";
$target = $target . $ran2.$ext;
if(move_uploaded_file($_FILES['uploaded']['tmp_name'], $target))
{
echo "The file has been uploaded as ".$ran2.$ext;
}
else
{
echo "Sorry, there was a problem uploading your file.";
}
?> 发布于 2012-11-07 16:19:24
我认为您的form标记上缺少enctype="multipart/form-data"属性。否则,表单将使用application/x-www-form-urlencoded的默认enctype提交,该默认not不支持二进制数据上传。
https://stackoverflow.com/questions/13041895
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