Ios 如何在UITextView中显示可单击链接
我试图在UITextview中显示带有可单击链接的属性字符串。我已经创建了一个简单的测试项目，看看哪里出了问题，但仍然无法找出答案。我尝试过启用用户交互和设置shouldInteractWithURLs委托方法，但仍然不起作用。这是我的代码（对于只包含textview的视图控制器） 下面是我实现的委托方法：Ios 如何在UITextView中显示可单击链接,ios,swift,uitextview,Ios,Swift,Uitextview,我试图在UITextview中显示带有可单击链接的属性字符串。我已经创建了一个简单的测试项目，看看哪里出了问题，但仍然无法找出答案。我尝试过启用用户交互和设置shouldInteractWithURLs委托方法，但仍然不起作用。这是我的代码（对于只包含textview的视图控制器） 下面是我实现的委托方法： func textViewShouldBeginEditing(textView: UITextView) -> Bool { return false } func tex
func textViewShouldBeginEditing(textView: UITextView) -> Bool {
return false
}
func textView(textView: UITextView, shouldInteractWithURL URL: NSURL, inRange characterRange: NSRange) -> Bool {
return true
}
这仍然不起作用。我已经搜索了这个主题，但没有任何帮助。非常感谢。只需在情节提要中选择
UITextViewselectable链接。如下图所示。确保未选中“可编辑”Swift 3 iOS 10:这里有一个可点击的扩展UITextView，只要链接以www.exmaple.com开头，它就会自动检测textview中的网站。例如：www.exmaple.com如果它存在于文本中的任何地方，则可以点击。下面是课程：
import Foundation
import UIKit
public class ClickableTextView:UITextView{
var tap:UITapGestureRecognizer!
override public init(frame: CGRect, textContainer: NSTextContainer?) {
super.init(frame: frame, textContainer: textContainer)
print("init")
setup()
}
required public init?(coder aDecoder: NSCoder) {
super.init(coder: aDecoder)
setup()
}
func setup(){
// Add tap gesture recognizer to Text View
tap = UITapGestureRecognizer(target: self, action: #selector(self.myMethodToHandleTap(sender:)))
// tap.delegate = self
self.addGestureRecognizer(tap)
}
func myMethodToHandleTap(sender: UITapGestureRecognizer){
let myTextView = sender.view as! UITextView
let layoutManager = myTextView.layoutManager
// location of tap in myTextView coordinates and taking the inset into account
var location = sender.location(in: myTextView)
location.x -= myTextView.textContainerInset.left;
location.y -= myTextView.textContainerInset.top;
// character index at tap location
let characterIndex = layoutManager.characterIndex(for: location, in: myTextView.textContainer, fractionOfDistanceBetweenInsertionPoints: nil)
// if index is valid then do something.
if characterIndex < myTextView.textStorage.length {
let orgString = myTextView.attributedText.string
//Find the WWW
var didFind = false
var count:Int = characterIndex
while(count > 2 && didFind == false){
let myRange = NSRange(location: count-1, length: 2)
let substring = (orgString as NSString).substring(with: myRange)
// print(substring,count)
if substring == " w" || (substring == "w." && count == 3){
didFind = true
// print("Did find",count)
var count2 = count
while(count2 < orgString.characters.count){
let myRange = NSRange(location: count2 - 1, length: 2)
let substring = (orgString as NSString).substring(with: myRange)
// print("Did 2",count2,substring)
count2 += 1
//If it was at the end of textView
if count2 == orgString.characters.count {
let length = orgString.characters.count - count
let myRange = NSRange(location: count, length: length)
let substring = (orgString as NSString).substring(with: myRange)
openLink(link: substring)
print("It's a Link",substring)
return
}
//If it's in the middle
if substring.hasSuffix(" "){
let length = count2 - count
let myRange = NSRange(location: count, length: length)
let substring = (orgString as NSString).substring(with: myRange)
openLink(link: substring)
print("It's a Link",substring)
return
}
}
return
}
if substring.hasPrefix(" "){
print("Not a link")
return
}
count -= 1
}
}
}
func openLink(link:String){
if let checkURL = URL(string: "http://\(link.replacingOccurrences(of: " ", with: ""))") {
if UIApplication.shared.canOpenURL(checkURL) {
UIApplication.shared.open(checkURL, options: [:], completionHandler: nil)
print("url successfully opened")
}
} else {
print("invalid url")
}
}
public override func didMoveToWindow() {
if self.window == nil{
self.removeGestureRecognizer(tap)
print("ClickableTextView View removed from")
}
}
}
override func viewDidLoad() {
super.viewDidLoad()
let linkAttributes = [
NSLinkAttributeName: NSURL(string: "http://stalwartitsolution.co.in/luminutri_flow/terms-condition")!
] as [String : Any]
let attributedString = NSMutableAttributedString(string: "Please tick box to confirm you agree to our Terms & Conditions, Privacy Policy, Disclaimer. ")
attributedString.setAttributes(linkAttributes, range: NSMakeRange(44, 18))
attributedString.addAttribute(NSUnderlineStyleAttributeName, value: NSNumber(value: 1), range: NSMakeRange(44, 18))
textview.delegate = self
textview.attributedText = attributedString
textview.linkTextAttributes = [NSForegroundColorAttributeName: UIColor.red]
textview.textColor = UIColor.white
}
func textView(_ textView: UITextView, shouldInteractWith URL: URL, in characterRange: NSRange, interaction: UITextItemInteraction) -> Bool {
return true
}
谢谢事实上，我只是在发帖后马上就想出来了（已经花了几个小时在这上面了，想想看）。我已经选中了可选链接和链接，但我需要取消选中“可编辑”以使其正常工作。Rashwan，如果我想使文本视图可编辑，但同时也希望在文本视图退出第一响应者时自动检测链接，如在iOS Notes应用程序中？我可以这样做吗？知道在这种情况下什么是链接吗？@GraysonM，根据文档，它是HTTP链接。在textview中，google.in不显示为链接。如果你有什么解决办法，请帮忙
override func viewDidLoad() {
super.viewDidLoad()
let linkAttributes = [
NSLinkAttributeName: NSURL(string: "http://stalwartitsolution.co.in/luminutri_flow/terms-condition")!
] as [String : Any]
let attributedString = NSMutableAttributedString(string: "Please tick box to confirm you agree to our Terms & Conditions, Privacy Policy, Disclaimer. ")
attributedString.setAttributes(linkAttributes, range: NSMakeRange(44, 18))
attributedString.addAttribute(NSUnderlineStyleAttributeName, value: NSNumber(value: 1), range: NSMakeRange(44, 18))
textview.delegate = self
textview.attributedText = attributedString
textview.linkTextAttributes = [NSForegroundColorAttributeName: UIColor.red]
textview.textColor = UIColor.white
}
func textView(_ textView: UITextView, shouldInteractWith URL: URL, in characterRange: NSRange, interaction: UITextItemInteraction) -> Bool {
return true
}