无法获取图像URL放入Firebase中的数据库

时间:2018-10-16 10:28:44

标签: ios swift firebase firebase-realtime-database firebase-storage

我在个人资料图片中添加了点击手势,该图片可以从iPhone拾取图片

let tapGesture = UITapGestureRecognizer(target: self, action: #selector(SignUpVC.handleSelectProfileImgView))
profileImage.isUserInteractionEnabled = true
profileImage.addGestureRecognizer(tapGesture)

func imagePickerController(_ picker: UIImagePickerController, didFinishPickingMediaWithInfo info: [UIImagePickerController.InfoKey : Any]) {
    if let pickedImage = info[UIImagePickerController.InfoKey.originalImage] as? UIImage {
        profileImage.image = pickedImage
        selectedImage = pickedImage
    }
    dismiss(animated: true, completion: nil)
}

此后,我曾尝试将此图片放入 STORAGE ,然后在数据库中获取downloadURL设置

func uploadProfileImage(userUID: String, completion: @escaping (_ url: URL?) -> Void) {
    let storageRef = Storage.storage().reference().child("profile_images").child(userUID)

    if let imageData = selectedImage.jpegData(compressionQuality: 0.75) {
        storageRef.putData(imageData, metadata: nil) { (metadata, error) in
            guard metadata != nil else { return }
            storageRef.downloadURL(completion: { (url, error) in
                guard let url = url else {return}
                completion(url)
            })
        }
    }
}


@IBAction func signUpPressed(_ sender: Any) {
    Auth.auth().createUser(withEmail: (emailTextField.text!), password: (passwordTextField.text!)) { (result, error) in
        if let _eror = error {
            print(_eror.localizedDescription )
        } else {
            let userUID = result?.user.uid
            self.uploadProfileImage(userUID: userUID!, completion: { (url) in
                guard let url = url else {return}
                Database.database().reference().child("users").child(userUID!).setValue(["username":self.usernameTextField.text!, "email":self.emailTextField.text!, "password":self.passwordTextField.text!, "profileImgURL":url])
            })
        }
    }
}

我试图不调用 uploadProfileImage()函数,而只是将值username,email和passsword设置为数据库,它可以工作。但是,当我调用该函数uploadProfileImage()并获取URL时,它将失败。

我检查了存储空间,发现图像已成功上传。这里的问题是我无法获取downloadURL。

请帮助我。非常感谢!

1 个答案:

答案 0 :(得分:0)

storageRef.putData(imageData, metadata: nil) { (metadata, error) in
    if error != nil {
        print("error")
    } else {
        // your uploaded photo url.
        storageRef.downloadURL(completion: { (url, error) in
            if error != nil {
                print("error")
            } else {
                print(url!)
            }
    )}
}

您可以使用此方法并成功完成downloadURL,可以获得下载URL,您可以在浏览器中浏览该URL,以显示上载的图像。