Я использую приведенный ниже код для загрузки изображения на мой сервер, код отправляет запрос, но ответ от API всегда (была ошибка).
В то же время загрузки изображения также некоторая информация об этом изображении будет сохранена в mySql.
Вот мой код, я использую Xcode 10.1 и Swift 4.2
@IBAction func uploadImage(_ sender: Any) {
self.showActivityIndicator()
//Post URL
let url = "https://website.com/folder/include/upload.php"
//Getting text from textFiled!
let name = nameField.text!
let age = ageField.text!
//Call Parameters
let params: Parameters = ["name": name,"age": age]
//Checking image place holder
let image = UIImage(named: "map.png")
//Checking if empty name or age fileds
if name.isEmpty || age.isEmpty{
self.hideActivityIndicator()
myAlert(title: "Error", msg: "Make sure you enter all the required information!")
}
//Checking if image is not selected!!
else if imageView.image == image
{
self.hideActivityIndicator()
myAlert(title: "Error", msg: "Make sure you choose an image!")
}else{
let imageToUpload = self.imageView.image!
Alamofire.upload(multipartFormData:
{
(multipartFormData) in
multipartFormData.append(imageToUpload.jpegData(compressionQuality: 0.75)!, withName: "image", fileName: self.generateBoundaryString(), mimeType: "image/jpeg")
for (key, value) in params
{
multipartFormData.append((value as AnyObject).data(using: String.Encoding.utf8.rawValue)!, withName: key)
}
}, to:url,headers:nil)
{ (result) in
switch result {
case .success(let upload,_,_ ):
upload.uploadProgress(closure: { (progress) in
//Print progress
self.showActivityIndicator()
})
upload.responseJSON
{ response in
//print response.result
if let result = response.result.value {
//Calling response from API
let message = (result as AnyObject).value(forKey: "message") as! String
let status = (result as AnyObject).value(forKey: "status") as! String
//Case Success
if status == "1" {
self.hideActivityIndicator()
print("Your Results are ====> ",result)
self.myAlert(title: "Data Upload", msg: message)
self.imageView.image = UIImage(named: "map.png")
self.nameField.text = ""
self.ageField.text = ""
}else{
self.hideActivityIndicator()
self.myAlert(title: "Error Uploading", msg: message)
}
}
}
case .failure(let encodingError):
print(encodingError)
break
}
}
}
}
}
Вот код файла PHP:
<?php
include 'include/connect.php';
//Get Param Data
$name = $_POST["name"];
$age = $_POST["age"];
$xName = mysqli_real_escape_string($conn, $name);
$xAge = mysqli_real_escape_string($conn, $age);
//Results Array
$result = array();
//Image setup
$uploads_dir = 'img';
$tmp_name = $_FILES["image"]["tmp_name"];
$image_name = basename($_FILES["image"]["name"]);
$supported_image = array('gif','jpg','jpeg','png');
$ext = strtolower(pathinfo($image_name, PATHINFO_EXTENSION));
if(empty($xName) || empty($xAge)|| empty($image_name))
{
// Send some dummy result back to the iOS app
$result["message"] = "Sorry, there was an error uploading your file.";
$result["status"] = "0";
$result["post"] = $_POST;
$result["files"] = $_FILES;
}
if (! in_array($ext, $supported_image))
{
// Send some dummy result back to the iOS app
$result["message"] = "Sorry, Image extension is not Allowed!";
$result["status"] = "0";
$result["post"] = $_POST;
$result["files"] = $_FILES;
}
else
{
$query ="INSERT INTO images (name, age, image) VALUES ('$xName', '$xAge','$image_name')";
if (mysqli_query($conn, $query)) {
move_uploaded_file($tmp_name,"$uploads_dir/$image_name");
// Send some dummy result back to the iOS app
$result["message"] = "Data has been uploaded successfully.";
$result["status"] = "1";
$result["post"] = $_POST;
$result["files"] = $_FILES;
}
}
echo json_encode($result);
?>
Кажется, что в ответе API отсутствует некоторая информация, но я заполняю два поля необходимой информацией (имя и возраст).
Мне не хватает того, что мне не хватает для завершения загрузки изображения и его информации.
Спасибо