我有一个UITextview
我想把几个单词作为链接,以便我可以检测到它们并获得回调函数。我已将UITextView
的链接属性设置为已检查,并在委托方法下实现。
func textView(_ textView: UITextView, shouldInteractWith URL: URL, in characterRange: NSRange) -> Bool {
print("hello")
return true
}
@available(iOS 10.0, *)
func textView(_ textView: UITextView, shouldInteractWith URL: URL, in characterRange: NSRange, interaction: UITextItemInteraction) -> Bool {
print("bye")
return true
}
我还使用NSMutableString
let str = NSMutableAttributedString(string: self.tvBottom.text as String, attributes: [NSFontAttributeName:UIFont.systemFont(ofSize: 17.0)])
str.addAttribute(NSLinkAttributeName, value: "http://www.google.com", range: NSRange(location: 4, length: 14))
str.addAttribute(NSForegroundColorAttributeName, value: UIColor.blue , range: NSRange(location: 4, length: 14))
let boldFontAttribute = [NSFontAttributeName: UIFont.boldSystemFont(ofSize: 17.0)]
str.addAttributes(boldFontAttribute, range: NSRange(location: 4, length: 14))
str.addAttribute(NSUnderlineStyleAttributeName , value: NSUnderlineStyle.styleSingle.rawValue, range: NSRange(location: 4, length: 14))
self.tvBottom.attributedText = str
我该怎么做?
答案 0 :(得分:0)
要检测特定行上的点击(如问题标题所示),则以下代码可以完成此任务:
func didTapTextView(recognizer: UITapGestureRecognizer) {
if recognizer.state == .recognized {
let location = recognizer.location(ofTouch: 0, in: textView)
if location.y >= 0 && location.y <= textView.contentSize.height {
guard let font = textView.font else {
return
}
let line = Int((location.y - textView.textContainerInset.top) / font.lineHeight) + 1
print("Line is \(line)")
}
}
}
以下代码用于在我们的textView上检测到点按时调用上述方法。
let tap = UITapGestureRecognizer(target: self, action: #selector(didTapTextView(recognizer:)))
tap.cancelsTouchesInView = false
textView.addGestureRecognizer(tap)
但是,如果您只想检测链接上的点击,那么您应该符合UITextViewDelegate
协议,并实现以下方法:
func textView(_ textView: UITextView, shouldInteractWith URL: URL, in characterRange: NSRange, interaction: UITextItemInteraction) -> Bool {
return true
}
当然,您需要将委托添加到textView(例如viewDidLoad()
)
textView.delegate = self
P.S。您的UITextView
不应该是可编辑的。