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我希望标签可点击以在 safari 中打开 url，点击时在电子邮件中打开电话号码和电子邮件地址

override func viewDidLoad() {
  super.viewDidLoad()

  // DetailsSV.contentSize.height=1120

  print("idis \(self.strUserid)")

  let ref = Firebase(url: "https://businesswallet.firebaseio.com/Users")

  ref.childByAppendingPath(self.strUserid as String).observeEventType(.Value, withBlock: { snapshot in
    if let dict = snapshot.value as? NSMutableDictionary{
      print("dict is \(dict)")
      if let Email = dict["Email"] as? String {
        self.EmailL.text = Email
      }
      if let name = dict["BusinessName"] as? String {
        self.BusinessNameL.text = name
        self.navigationItem.title = name
      }
      if let ShortDescription = dict["ShortDescription"] as? String {
        self.ShortDescriptionL.text = ShortDescription
      }
      if let City = dict["City"] as? String {
        self.CityL.text = City
      }
      if let ContactMe = dict["ContactMe"] as? String {
        self.ContactMeL.text = ContactMe
      }
      if let PhoneNumber = dict["PhoneNumber"] as? String {
        self.PhoneNumberL.text = PhoneNumber
      }
      if let Website1 = dict["Website1"] as? String {
        self.Website1L.text = Website1
      }
      if let Website2 = dict["Website2"] as? String {
        self.Website2L.text = Website2
      }
      if let Category = dict["Category"] as? String {
        self.CategoryL.text = Category
      }
      if let Details = dict["Details"] as? String {
        self.DetailsTV.text = Details
      }
    }
  })

我添加了这个功能，但它不起作用：

if let Website1 = dict["Website1"] as? String {
  self.Website1L.text = Website1
  let weblurl = NSURL(string: "open page:\(Website1)")
  UIApplication.sharedApplication().openURL(weblurl!)
}
4

1 回答 1

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尝试使用 UITextView。如果您不希望用户能够编辑 UITextView 内的文本，您可以在 UITextView 的属性检查器中取消选中“可编辑”。如果您希望 UITextView 能够检测链接、电话号码、事件和地址等文本字符串，您可以在 UITextView 的 Attributes Inspector 中检查适用的字符串。

如果您知道只想显示一个链接，只需添加一个带有您想要的文本的按钮，然后让它将用户发送到 Safari。

于 2016-05-02T19:36:52.483 回答