Question

我想发布照片并在Alphafire中快速使用php 这是我在php中的代码

if(isset($_POST)){
    $user_member = filter_var($_POST['member'] , FILTER_SANITIZE_STRING);
    $usersModel = new UsersModel();
    if($usersModel->checkUserMemberId($user_member)){
        $title = filter_var($_POST['title'] , FILTER_SANITIZE_STRING);
        $desc  = filter_var($_POST['desc'] , FILTER_SANITIZE_STRING);
        $image = isset($_FILES['file']['name']) ? $_FILES['file'] : FALSE;
        $Tickits_model = new Tickets_Model();
        if($Tickits_model->addNewTickit($title, $desc, $image , $user_member)){
            header('HTTP/1.1 200');
            header('Content-Type: application/json; charset=UTF-8');
            die(json_encode('success'));
        }else{
            header('HTTP/1.1 400 Bad Request');
            header('Content-Type: application/json; charset=UTF-8');
            die(json_encode('error'));
        }
    }
}

这是我的快速代码

func uploadimage() {

    var myUrl = "myurl"
    myUrl = myUrl.addingPercentEncoding(withAllowedCharacters: .urlFragmentAllowed)!

    let url = URL(string: myUrl)

    let image = isimage.image!
    guard let data = image.jpegData(compressionQuality: 0.75) else {
        return
    }

    Alamofire.upload(multipartFormData: { (form) in
        form.append(data, withName: "image" ( i try using "file" and same problem) , fileName: "file.jpg", mimeType: "image/jpg")
    }, to: url! , encodingCompletion: { result in
        switch result {
        case .success(let upload, _, _):
            upload.responseString
            {
                response in
                print(response)
                print(result)
                print(response.description)
            }
        case .failure(let encodingError):
            print(encodingError)
        }
    })
}

这是显示的错误

success(request: 2019-01-12 18:09:27.000943+0300 NFSTORE[6346:2103776] CredStore - performQuery - Error copying matching creds.  Error=-25300, query={
class = inet;
"m_Limit" = "m_LimitAll";
ptcl = htps;
"r_Attributes" = 1;
sdmn = "myurl";
srvr = "myurl";
sync = syna;

我尝试将“图像”更改为“文件”，并遇到相同的问题。我使用Alamofire中的另一种方法和一种上传图像的方法发送其他参数（title-desc-member）。我不知道这是否正确，但是如果不能给出最好的主意，请

使用Alamofire快速发布照片

0 个答案: