Php 无法将带有Alamofire的图像上载到我的服务器
我正在尝试将一个图像上载到我的服务器,但我的代码无效。以下是我的swift代码:Php 无法将带有Alamofire的图像上载到我的服务器,php,ios,swift,alamofire,backend,Php,Ios,Swift,Alamofire,Backend,我正在尝试将一个图像上载到我的服务器,但我的代码无效。以下是我的swift代码: import UIKit import Alamofire import SwiftyJSON class ViewController: UIViewController { @IBOutlet weak var theOneImage: UIImageView! let url = "http://xxx.xxx.xxx.xxx/Twitter/UploadImage.php" let avaImage =
import UIKit
import Alamofire
import SwiftyJSON
class ViewController: UIViewController {
@IBOutlet weak var theOneImage: UIImageView!
let url = "http://xxx.xxx.xxx.xxx/Twitter/UploadImage.php"
let avaImage = UIImage(named: "cupid")
override func viewDidLoad() {
super.viewDidLoad()
theOneImage.image = avaImage
uploadImage(image: avaImage)
}
func uploadImage(image : UIImage?){
let imageData = avaImage?.pngData()
Alamofire.upload(multipartFormData: { (MultipartFormData) in
MultipartFormData.append(imageData!, withName: "image")
}, to: url) { (encodingResult) in
switch encodingResult{
case .success(let upload, _, _):
upload.responseJSON{ response in
print(response.request!) // original URL request
print(response.response!) // URL response
print(response.data!) // server data
print(response.result) // result of response serialization
if let JSON = response.result.value {
print("JSON: \(JSON)")
}
}
case .failure(let encodingError):
print(encodingError)
}
}
}
}
以下是我的php代码:
if (empty($_FILES["image"])){
$response = array("error" => "no data");
}else{
$response["error"] = "NULL";
$filename = uniqid() . ".jpg";
if(move_uploaded_file($_FILES['image']['test'],'/ava')){
$response["status"] = "success";
$response["filepath"] = "http:" . $filename;
$response["filename"] = "" . $_FILES["file"]["name"];
}else{
$response['status'] = "Failure";
$response['error'] = "".$_FILES["image"]["error"];
$response['name'] = "".$_FILES["image"]["name"];
$response['path'] = "".$_FILES["image"]["tmp_name"];
$response['type'] = "".$_FILES["image"]["type"];
$response['size'] = "".$_FILES["image"]["size"];
}
}
echo json_encode($response);
我得到的回应是,在UploadImage.php中,$\u文件[“image”]是空的
以下是我的服务器目录:
请帮忙。谢谢大家! 404通常表示“找不到”,您是否正在上载到正确的端点?哦,谢谢,我忘了包含.php。这次我得到了一个JSON,但它显示“没有数据”。根据我的php文件,这意味着$_文件[“image”]是空的。我们确定后端工作正常吗?我对PHP没有任何线索,但我可能知道SwiftI有什么问题,可以成功发送请求,但问题是后端没有接收到我的图像。swift似乎没有正确发送数据可能是