UIButton无法响应点击

问题描述 投票:0回答:1

帮帮我。当我点击时我的按钮不起作用,我尝试从堆栈溢出的另一篇文章中尝试其他建议。但没有一个起作用,这是我的代码。

    let saveBtn             = GActionBtn(type: .system)

    lazy var footer: UIView = {
        let v = UIView()
        v.isUserInteractionEnabled = true
        v.addSubview(saveBtn)
        saveBtn.isUserInteractionEnabled = true
        saveBtn.addTarget(self, action: #selector(handleSave), for: .touchUpInside)
        saveBtn.backgroundColor = #colorLiteral(red: 0.4470588235, green: 0.6274509804, blue: 0.3960784314, alpha: 1)
        saveBtn.layer.cornerRadius = 27.5
        let attributeString = NSAttributedString(string: "Save".localized(), attributes: [NSAttributedString.Key.font: UIFont(name: "NunitoSans-Bold", size: 16), NSAttributedString.Key.foregroundColor: UIColor.white])
        saveBtn.setAttributedTitle(attributeString, for: .normal)
        NSLayoutConstraint.activate([
            saveBtn.centerXAnchor.constraint(equalTo: v.centerXAnchor),
            saveBtn.centerYAnchor.constraint(equalTo: v.centerYAnchor, constant: 54),
            saveBtn.widthAnchor.constraint(equalToConstant: 312),
            saveBtn.heightAnchor.constraint(equalToConstant: 52)
        ])
        return v
    }()

    @objc private func handleSave() {
        print("save")
    }

    private func setupTableView() {
        tableView = UITableView(frame: .zero, style: .grouped)
        tableView.contentInset = UIEdgeInsets(top: 0, left: 0, bottom: 80, right: 0)
        tableView.separatorStyle = .none
        tableView.keyboardDismissMode = .interactive
        tableView.backgroundColor = .white
        tableView.tableFooterView = footer
    }
ios swift uibutton
1个回答
-1
投票

设置非零帧/约束为

 let v = UIView()
 tableView = UITableView(frame: .zero, style: .grouped)
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