Swift 3 打开链接

Question

我正在尝试在点击按钮时运行此功能：

@IBAction func openLink(_ sender: UIButton) {
    let link1 = "https://www.google.com/#q="
    let link2 = birdName.text!
    let link3 = link2.replacingOccurrences(of: " ", with: "+") //EDIT
    let link4 = link1+link3
    guard
        let query = link4.addingPercentEncoding( withAllowedCharacters: .urlQueryAllowed),
        let url = NSURL(string: "https://google.com/#q=\(query)")
        else { return }
    UIApplication.shared.openURL(URL(url))
}

但是，最后一行被标记为“不能调用非函数类型“UIApplication”的值。这个语法来自这里，所以我不确定发生了什么。

Answer 1

使用 guard 解开 textfield 文本属性，替换出现的地方，在结果中添加百分比编码并从结果字符串创建 URL：

试试这样：

guard
    let text = birdName.text?.replacingOccurrences(of: " ", with: "+"),
    let query = text.addingPercentEncoding(withAllowedCharacters: .urlQueryAllowed),
    let url = URL(string: "https://google.com/#q=" + query)
else { return }
if #available(iOS 10.0, *) {
    UIApplication.shared.open(url)
} else {
    UIApplication.shared.openURL(url)
}