相关疑难解决方法(0)

我只是在Swift imagePickerController中没有"info"所以我不知道如何获取url并将其转换为数据以发送到Web服务.

func imagePickerController(picker: UIImagePickerController, didFinishPickingMediaWithInfo info: [NSObject : AnyObject]) {

  var videoDataURL = info[UIImagePickerControllerMediaURL] as! NSURL!
  var videoFileURL = videoDataURL.filePathURL
  var video = NSData.dataWithContentsOfMappedFile("\(videoDataURL)")
}

我正在做一个项目Swift 3- xcode 8,我正在尝试使用核心数据来保存并在数据库表"用户"中显示一些图像.此图片是其个人资料中的用户照片.

现在我已经设法保存字符串并从核心数据中显示它们,但我在处理图像时遇到了问题.

这是我到目前为止:

将USERS添加到核心数据中

func addUser() {
    let app = UIApplication.shared.delegate as! AppDelegate
    let context = app.persistentContainer.viewContext
    let request = NSFetchRequest<NSFetchRequestResult>(entityName: "Users")
    request.returnsObjectsAsFaults = false

    let newUser = NSEntityDescription.insertNewObject(forEntityName: "Users", into: context)

    if (firstName.text == "" && lastName.text == "" && contact.text == "" && email.text == "") { //if we have a user profile delete it
        deleteUser()
    } else { // add a new user profile
        newUser.setValue(firstName.text, forKey: "firstName")
        newUser.setValue(lastName.text, forKey: "lastName")
        newUser.setValue(contact.text, …