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我已经创建了一个API,所以android应用程序可以连接到网站我面对的问题是我无法提供访问权限以便将图像上传到服务器,至此我已创建所以图像将作为base64代码保存,但主要问题是,当每个API调用和图像上传它给我一个图像错误,没有找到无法加载文件流。下面是我得到用于上传图像的API不起作用
<b>Warning</b>: fopen(uploads/testimg.png): failed to open stream: No such
file or directory in <b>/home/begazed/public_html/api/classfiles/Photo.php</b> on line <b>87</b><br />
<br />
<b>Warning</b>: fwrite() expects parameter 1 to be resource, boolean given in <b>/home/begazed/public_html/api/classfiles/Photo.php</b> on line <b>88</b><br />
<br />
<b>Warning</b>: fclose() expects parameter 1 to be resource, boolean given in <b>/home/begazed/public_html/api/classfiles/Photo.php</b> on line <b>89</b><br />
{"result":1,"message":"Image upload complete, Please check your php file directory"}
这个错误是我创建上传图像到服务器的功能
function upload_photo($connect, $base, $filename) {
$binary = base64_decode($base);
$query = $connect->prepare("INSERT INTO test_photo (filename) VALUES (:filename)");
$query->execute(array(':filename' => $filename));
header('Content-Type: bitmap; charset=utf-8');
$file = fopen('uploads/'.$filename, 'wb');
fwrite($file, $binary);
fclose($file);
$err['result'] = 1;
$err['message'] = 'Image upload complete, Please check your php file directory';
return json_encode($err);
}
从那里请求被处理了主要的API文件
switch($action)
{
case 'upload_photo':
$base = $_REQUEST['image'];
$filename = $_REQUEST['filename'];
echo $retval = $photo->upload_photo($connect, $base, $filename);
break;
default:
echo "Invalid Request";
break;
}
tthis是用于拨打电话的网址
www.domainname.com/api/api.php?action=upload_photo & fimename =(文件名)&图像= iamge NAE
谁能帮我如果你的PHP的来解决这个错误
没有人可以帮助我:-( –