Question

伙计们，我有使用swift3通过alamofire 4发布json数据的问题，以及使用php在XAMPP服务器端检索json数据的问题。我的swift 3代码确实触发了XAMPP的PHP脚本，但不知怎的，我无法通过php中的$ _POST变量获取它这是我的代码，

func uploadImage(image: UIImage){
    //Now use image to create into NSData format
    let imageData:NSData = UIImagePNGRepresentation(image)! as NSData      
    //convert the nsdata to base64 encoded string
       let strBase64:String = imageData.base64EncodedString(options: .lineLength64Characters)
   // let parameters = ["image": strBase64] as Dictionary
    let parameters = ["image": strBase64]      
    print(strBase64) 
  Alamofire.request("http://localhost/Test/api/UploadPhoto.php",method: .post, parameters: parameters, encoding: JSONEncoding.default).response { response in            
    print(response)                   
                }   
}

这是我的服务器端代码（脚本确实是由来自alamofire的调用触发的，但不知怎的，我只能通过调用$ _POST [＆＃34; image＆＃34;]来获取数据）

    <?php
//scripts below did get triggered, but can't get the json data through          calling $_POST["image"];
$imageString = $_POST["image"];
$filename_path = md5(time().uniqid()).".png"; 
$data = base64_decode($imageString);
file_put_contents('../AllImages/'.$filename_path, $data);
echo json_encode($_POST["image"]);
?>

如果可能的话，请帮助我，我已经挣扎了近一个星期，但却找不到很多线索谢谢

Answer 1

我找到了解决这个问题的方法，基本上，我使用urlsession.shared.datatask帮助我而不是alamofire发布请求，这是我的ios侧码

     func uploadImage(image: UIImage, completionHandler: @escaping (String) ->()){

   // Now use image to create into NSData format
            let imageData:NSData = UIImagePNGRepresentation(image)! as NSData

            //convert the nsdata to base64 encoded string

               let strBase64:String = imageData.base64EncodedString(options: .lineLength64Characters)
    // prepare json data
    let json: [String: Any] = ["image": strBase64]

    let jsonData = try? JSONSerialization.data(withJSONObject: json)

    // create post request
    let url = URL(string: "http://10.10.10.72/Test/api/UploadPhoto.php")!
    var request = URLRequest(url: url)
    request.httpMethod = "POST"

    // insert json data to the request
    request.httpBody = jsonData

    let task = URLSession.shared.dataTask(with: request) { data, response, error in

        do {
            guard let data = data else {
                throw JSONError.NoData
            }
            guard let json = try JSONSerialization.jsonObject(with: data, options: []) as? [String: AnyObject] else {
                throw JSONError.ConversionFailed
            }

             completionHandler(json["sign"] as! String)

        } catch let error as JSONError {
            print(error.rawValue)
        } catch let error as NSError {
            print(error.debugDescription)
        }

        }
    task.resume()
}

我使用字典来存储我的数据，并将其转换为json数据格式以发送到服务器

     let json: [String: Any] = ["image": strBase64]
     let jsonData = try? JSONSerialization.data(withJSONObject: json)

然后在php端，我使用

检索它

    $entityBody = file_get_contents('php://input');

，然后我从json解码，它生成了一个数组，我可以通过引用图像访问我的值，所以完整的PHP代码如下：

    <?php
//get the posted json data
 $entityBody = file_get_contents('php://input');
//decode the json data
$decoded = json_decode($entityBody, TRUE);
$imageString = $decoded["image"];
//create a unique name for the image
 $filename_path = md5(time().uniqid()).".png"; 
//converted the image string back to image
 $data = base64_decode($imageString);
//put it on the desired location
file_put_contents('../AllImages/uploads/signature/'.$filename_path, $data);
 $response = array();
 //create the response 
 $response['sign'] = '../AllImages/uploads/signature/'.$filename_path;
 echo json_encode($response);
 ?>

请注意这里，我再次编码json数据作为从php到我的ios端的响应发送回来，你需要解码来自json的响应，所以完整的想法是如果你将值编码为json来自一方面，你需要从另一侧解码它才能正确访问价值，如果我错了就纠正我，我很高兴我的应用程序现在启动并运行所有请求：D

从alamofire swift3的post请求中通过$ _POST检索json数据？

1 个答案: