xcode Swift 3 打开链接
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Swift 3 Open Link
提问by Alex
I'm trying to run this function when a button is tapped:
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))
}
However, the last line is flagged as "cannot call value of non-function type "UIApplication". This syntax is from here, so I'm not sure whats going on.
但是，最后一行被标记为“无法调用非函数类型“UIApplication”的值。此语法来自此处，因此我不确定发生了什么。
回答by Leo Dabus
Use guard to unwrap the textfield text property, replacing the occurrences, add percent encoding to the result and create an URL from the resulting string:
使用 guard 来解开 textfield 文本属性，替换出现的内容，向结果添加百分比编码并从结果字符串创建一个 URL：
Try like this:
像这样尝试：
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)
}
