“Ustaw przycisk programowo szybki” Kod odpowiedzi

Ustaw przycisk programowo szybki

        let button = self.makeButton(title: "Login", titleColor: .blue, font: UIFont.init(name: "Arial", size: 18.0), background: .white, cornerRadius: 3.0, borderWidth: 2, borderColor: .black)
        view.addSubview(button)
        // Adding Constraints
        button.heightAnchor.constraint(equalToConstant: 40).isActive = true
        button.leftAnchor.constraint(equalTo: view.leftAnchor, constant: 40).isActive = true
        button.rightAnchor.constraint(equalTo: view.rightAnchor, constant: -40).isActive = true
        button.bottomAnchor.constraint(equalTo: view.bottomAnchor, constant: -400).isActive = true
        button.addTarget(self, action: #selector(pressed(_ :)), for: .touchUpInside)

       // Define commmon method
        func makeButton(title: String? = nil,
                           titleColor: UIColor = .black,
                           font: UIFont? = nil,
                           background: UIColor = .clear,
                           cornerRadius: CGFloat = 0,
                           borderWidth: CGFloat = 0,
                           borderColor: UIColor = .clear) -> UIButton {
        let button = UIButton()
        button.translatesAutoresizingMaskIntoConstraints = false
        button.setTitle(title, for: .normal)
        button.backgroundColor = background
        button.setTitleColor(titleColor, for: .normal)
        button.titleLabel?.font = font
        button.layer.cornerRadius = 6.0
        button.layer.borderWidth = 2
        button.layer.borderColor = UIColor.red.cgColor
        return button
      }
        // Button Action
         @objc func pressed(_ sender: UIButton) {
                print("Pressed")
          }