我使用下面的代码来访问html 5摄像头并将图片上传到服务器。HTML 5使用php的摄像头访问和上传文件
HTML代码
<form action="upload.php" method="post" enctype="multipart/form-data">
<input type="file" name="image" accept="image/*" capture>
<input type="submit" value="Upload">
</form>
upload.php的代码
<?php
$target_path = "upload/";
$target_path = $target_path . basename($_FILES['uploadedfile']['name']);
if(move_uploaded_file($_FILES['uploadedfile']['tmp_name'], $target_path)) {
echo "The file ". basename($_FILES['uploadedfile']['name']).
" has been uploaded";
} else{
echo "There was an error uploading the file, please try again!";
}
?>
问题是,当我测试它显示的代码“有一个上传文件时出错,请重试!” 。任何人都可以帮我弄清楚问题出在哪里?
下面的代码适合我。
HTML代码:
<form enctype="multipart/form-data" action="upload.php" method="POST">
<input type="hidden" name="MAX_FILE_SIZE" value="100000" />
Choose a file to upload: <input name="uploadedfile" type="file" /><br />
<input type="submit" value="Upload File" />
</form>
PHP代码是与上述相同。
在此先感谢。
'move_uploaded_file()'的输出是什么?您可能无权将文件放入目标文件夹或发生其他错误。检查你的错误日志。 – chrki 2013-02-17 11:51:00