HTML 5使用php访问和上传文件

Question

我使用下面的代码进行html 5相机访问并将图像上传到服务器.

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!";
}
?>

问题是当我测试代码时,它显示"上传文件时出错,请再试一次!" .任何人都可以帮我弄明白问题出在哪里？

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代码与上面相同.

Answer 1

在第一种情况下,当第二种情况被命名为"uploadedfile"时,您的文件输入名称被命名为"image".你的PHP期待'uploadedfile'.

要解决此问题,您需要将代码(第一种情况)更改为:

<form action="upload.php" method="post" enctype="multipart/form-data">  
   <input type="file" name="uploadedfile" accept="image/*" capture>  
   <input type="submit" value="Upload">  
</form>

Answer 2

如果你想使用HTML5和设备的相机,有一个HTML5功能:例如:

<input type="file" accept="image/*" capture="camera" />

从这里:https://coderwall.com/p/epwmoa

它正在开发适用于Android的Chrome(使用三星Galaxy S2).