【发布时间】:2015-01-07 15:52:36
【问题描述】:
嗨,我有这个代码 uploader.php:
class Uploader
{
var $filePath;
var $uploadURL;
var $formFileVariableName;
var $postParams = array ();
function Uploader($filePath, $uploadURL, $formFileVariableName, $otherParams = false)
{
$this->filePath = $filePath;
$this->uploadURL = $uploadURL;
if(is_array($otherParams) && $otherParams != false)
{
foreach($otherParams as $fieldName => $fieldValue)
{
$this->postParams[$fieldName] = $fieldValue;
}
}
$this->postParams[$formFileVariableName] = "@" . $filePath;
}
function UploadFile()
{
$ch = curl_init();
curl_setopt($ch, CURLOPT_URL, $this->uploadURL);
curl_setopt($ch, CURLOPT_POST, 1);
curl_setopt($ch, CURLOPT_POSTFIELDS, $this->postParams);
curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
$postResult = curl_exec($ch);
if (curl_errno($ch))
{
print curl_error($ch);
print "Unable to upload file.";
exit();
}
curl_close($ch);
return $postResult;
}
}
我是这样使用的:
require('uploader.php');
$upload_server = "http://developers.mytech.com.mx/files/api/upload.php";
$file = $_FILES['archivo']['tmp_name'];
$archivo = $file;
$extension = pathinfo($archivo, PATHINFO_EXTENSION);
$nombre_base = basename($archivo, '.'.$extension);
$upload = new Uploader($file, $upload_server,'file', array('api_key' => 'ic00n6yokd'));
$result = $upload->UploadFile();
if(preg_match("/upload_failed/", $result)){ echo "Upload failed."; }
if(preg_match("/error_tamano/", $result)){ echo "Archivo muy grande."; }
if(preg_match("/api_invalida/", $result)){ echo "Invalid API Code."; }
if(preg_match("/error_type/", $result)){ echo "Archivo no valido."; }
if(preg_match("/upload_success/", $result)){ echo $result; }
但问题是上传的是 tmp_name 作为文件而不是图像 但如果我改变这一行: $file = $_FILES['archivo']['tmp_name'];到 $file = $_FILES['archivo']['name'];
我得到这个错误:创建表单数据失败
我的问题是如何正确使用课程,以便我可以上传图片
【问题讨论】:
标签: php