许多应用程序都有文本,文本中是圆角矩形的web超链接,当我点击它们时,UIWebView就会打开。让我困惑的是,他们经常有自定义链接,例如,如果单词以#开头,它也是可点击的,应用程序通过打开另一个视图来响应。我该怎么做呢?是否可以用UILabel或者我需要UITextView或者其他什么?


当前回答

在Swift 3中工作,将整个代码粘贴在这里

    //****Make sure the textview 'Selectable' = checked, and 'Editable = Unchecked'

import UIKit

class ViewController: UIViewController, UITextViewDelegate {

    @IBOutlet var theNewTextView: UITextView!
    override func viewDidLoad() {
        super.viewDidLoad()

        //****textview = Selectable = checked, and Editable = Unchecked

        theNewTextView.delegate = self

        let theString = NSMutableAttributedString(string: "Agree to Terms")
        let theRange = theString.mutableString.range(of: "Terms")

        theString.addAttribute(NSLinkAttributeName, value: "ContactUs://", range: theRange)

        let theAttribute = [NSForegroundColorAttributeName: UIColor.blue, NSUnderlineStyleAttributeName: NSUnderlineStyle.styleSingle.rawValue] as [String : Any]

        theNewTextView.linkTextAttributes = theAttribute

     theNewTextView.attributedText = theString             

theString.setAttributes(theAttribute, range: theRange)

    }

    func textView(_ textView: UITextView, shouldInteractWith URL: URL, in characterRange: NSRange, interaction: UITextItemInteraction) -> Bool {

        if (URL.scheme?.hasPrefix("ContactUs://"))! {

            return false //interaction not allowed
        }

        //*** Set storyboard id same as VC name
        self.navigationController!.pushViewController((self.storyboard?.instantiateViewController(withIdentifier: "TheLastViewController"))! as UIViewController, animated: true)

        return true
    }

}

其他回答

    NSString *string = name;
    NSError *error = NULL;
    NSDataDetector *detector =
    [NSDataDetector dataDetectorWithTypes:(NSTextCheckingTypes)NSTextCheckingTypeLink | NSTextCheckingTypePhoneNumber
                                    error:&error];
    NSArray *matches = [detector matchesInString:string
                                         options:0
                                           range:NSMakeRange(0, [string length])];
    for (NSTextCheckingResult *match in matches)
    {
        if (([match resultType] == NSTextCheckingTypePhoneNumber))
        {
            NSString *phoneNumber = [match phoneNumber];
            NSLog(@" Phone Number is :%@",phoneNumber);
            label.enabledTextCheckingTypes = NSTextCheckingTypePhoneNumber;
        }

        if(([match resultType] == NSTextCheckingTypeLink))
        {
            NSURL *email = [match URL];
            NSLog(@"Email is  :%@",email);
            label.enabledTextCheckingTypes = NSTextCheckingTypeLink;
        }

        if (([match resultType] == NSTextCheckingTypeLink))
        {
            NSURL *url = [match URL];
            NSLog(@"URL is  :%@",url);
            label.enabledTextCheckingTypes = NSTextCheckingTypeLink;
        }
    }

    label.text =name;
}

以下是基于@Luca Davanzo的回答,重写touchesBegan事件而不是轻触手势:

import UIKit

public protocol TapableLabelDelegate: NSObjectProtocol {
   func tapableLabel(_ label: TapableLabel, didTapUrl url: String, atRange range: NSRange)
}

public class TapableLabel: UILabel {

private var links: [String: NSRange] = [:]
private(set) var layoutManager = NSLayoutManager()
private(set) var textContainer = NSTextContainer(size: CGSize.zero)
private(set) var textStorage = NSTextStorage() {
    didSet {
        textStorage.addLayoutManager(layoutManager)
    }
}

public weak var delegate: TapableLabelDelegate?

public override var attributedText: NSAttributedString? {
    didSet {
        if let attributedText = attributedText {
            textStorage = NSTextStorage(attributedString: attributedText)
        } else {
            textStorage = NSTextStorage()
            links = [:]
        }
    }
}

public override var lineBreakMode: NSLineBreakMode {
    didSet {
        textContainer.lineBreakMode = lineBreakMode
    }
}

public override var numberOfLines: Int {
    didSet {
        textContainer.maximumNumberOfLines = numberOfLines
    }
}


public override init(frame: CGRect) {
    super.init(frame: frame)
    setup()
}

public required init?(coder aDecoder: NSCoder) {
    super.init(coder: aDecoder)
    setup()
}

public override func layoutSubviews() {
    super.layoutSubviews()
    textContainer.size = bounds.size
}


/// addLinks
///
/// - Parameters:
///   - text: text of link
///   - url: link url string
public func addLink(_ text: String, withURL url: String) {
    guard let theText = attributedText?.string as? NSString else {
        return
    }

    let range = theText.range(of: text)

    guard range.location !=  NSNotFound else {
        return
    }

    links[url] = range
}

private func setup() {
    isUserInteractionEnabled = true
    layoutManager.addTextContainer(textContainer)
    textContainer.lineFragmentPadding = 0
    textContainer.lineBreakMode = lineBreakMode
    textContainer.maximumNumberOfLines  = numberOfLines
}

public override func touchesBegan(_ touches: Set<UITouch>, with event: UIEvent?) {
    guard let locationOfTouch = touches.first?.location(in: self) else {
        return
    }

    textContainer.size = bounds.size
    let indexOfCharacter = layoutManager.glyphIndex(for: locationOfTouch, in: textContainer)

    for (urlString, range) in links {
        if NSLocationInRange(indexOfCharacter, range), let url = URL(string: urlString) {
            delegate?.tapableLabel(self, didTapUrl: urlString, atRange: range)
        }
    }
}}

Drop-in解决方案作为UILabel上的一个类别(这假设你的UILabel使用一个带有NSLinkAttributeName属性的带属性字符串):

@implementation UILabel (Support)

- (BOOL)openTappedLinkAtLocation:(CGPoint)location {
  CGSize labelSize = self.bounds.size;

  NSTextContainer* textContainer = [[NSTextContainer alloc] initWithSize:CGSizeZero];
  textContainer.lineFragmentPadding = 0.0;
  textContainer.lineBreakMode = self.lineBreakMode;
  textContainer.maximumNumberOfLines = self.numberOfLines;
  textContainer.size = labelSize;

  NSLayoutManager* layoutManager = [[NSLayoutManager alloc] init];
  [layoutManager addTextContainer:textContainer];

  NSTextStorage* textStorage = [[NSTextStorage alloc] initWithAttributedString:self.attributedText];
  [textStorage addAttribute:NSFontAttributeName value:self.font range:NSMakeRange(0, textStorage.length)];
  [textStorage addLayoutManager:layoutManager];

  CGRect textBoundingBox = [layoutManager usedRectForTextContainer:textContainer];
  CGPoint textContainerOffset = CGPointMake((labelSize.width - textBoundingBox.size.width) * 0.5 - textBoundingBox.origin.x,
                                            (labelSize.height - textBoundingBox.size.height) * 0.5 - textBoundingBox.origin.y);
  CGPoint locationOfTouchInTextContainer = CGPointMake(location.x - textContainerOffset.x, location.y - textContainerOffset.y);
  NSInteger indexOfCharacter = [layoutManager characterIndexForPoint:locationOfTouchInTextContainer inTextContainer:textContainer fractionOfDistanceBetweenInsertionPoints:nullptr];
  if (indexOfCharacter >= 0) {
    NSURL* url = [textStorage attribute:NSLinkAttributeName atIndex:indexOfCharacter effectiveRange:nullptr];
    if (url) {
      [[UIApplication sharedApplication] openURL:url];
      return YES;
    }
  }
  return NO;
}

@end

在Swift 3中工作,将整个代码粘贴在这里

    //****Make sure the textview 'Selectable' = checked, and 'Editable = Unchecked'

import UIKit

class ViewController: UIViewController, UITextViewDelegate {

    @IBOutlet var theNewTextView: UITextView!
    override func viewDidLoad() {
        super.viewDidLoad()

        //****textview = Selectable = checked, and Editable = Unchecked

        theNewTextView.delegate = self

        let theString = NSMutableAttributedString(string: "Agree to Terms")
        let theRange = theString.mutableString.range(of: "Terms")

        theString.addAttribute(NSLinkAttributeName, value: "ContactUs://", range: theRange)

        let theAttribute = [NSForegroundColorAttributeName: UIColor.blue, NSUnderlineStyleAttributeName: NSUnderlineStyle.styleSingle.rawValue] as [String : Any]

        theNewTextView.linkTextAttributes = theAttribute

     theNewTextView.attributedText = theString             

theString.setAttributes(theAttribute, range: theRange)

    }

    func textView(_ textView: UITextView, shouldInteractWith URL: URL, in characterRange: NSRange, interaction: UITextItemInteraction) -> Bool {

        if (URL.scheme?.hasPrefix("ContactUs://"))! {

            return false //interaction not allowed
        }

        //*** Set storyboard id same as VC name
        self.navigationController!.pushViewController((self.storyboard?.instantiateViewController(withIdentifier: "TheLastViewController"))! as UIViewController, animated: true)

        return true
    }

}

老问题,但如果任何人都可以使用UITextView而不是UILabel,那就很容易了。标准网址,电话号码等将自动检测(并可点击)。

然而,如果你需要自定义检测,也就是说,如果你想在用户点击一个特定的单词后能够调用任何自定义方法,你需要使用NSAttributedStrings和一个NSLinkAttributeName属性,它将指向一个自定义URL方案(而不是在默认情况下使用http URL方案)。雷·温德里奇在这里报道

引用上述链接中的代码:

NSMutableAttributedString *attributedString = [[NSMutableAttributedString alloc] initWithString:@"This is an example by @marcelofabri_"];
[attributedString addAttribute:NSLinkAttributeName
                     value:@"username://marcelofabri_"
                     range:[[attributedString string] rangeOfString:@"@marcelofabri_"]];

NSDictionary *linkAttributes = @{NSForegroundColorAttributeName: [UIColor greenColor],
                             NSUnderlineColorAttributeName: [UIColor lightGrayColor],
                             NSUnderlineStyleAttributeName: @(NSUnderlinePatternSolid)};

// assume that textView is a UITextView previously created (either by code or Interface Builder)
textView.linkTextAttributes = linkAttributes; // customizes the appearance of links
textView.attributedText = attributedString;
textView.delegate = self;

要检测这些链接点击,实现这个:

- (BOOL)textView:(UITextView *)textView shouldInteractWithURL:(NSURL *)URL inRange:(NSRange)characterRange {
    if ([[URL scheme] isEqualToString:@"username"]) {
        NSString *username = [URL host]; 
        // do something with this username
        // ...
        return NO;
    }
    return YES; // let the system open this URL
}

PS:确保你的UITextView是可选的。