ios 如何使用 Swift 创建“超链接”？
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How can I create a "hyperlink" with Swift?
提问by AstroCB
I'm trying to make separate pieces of text UILabels clickable. What I'm looking for is commonly known as a hyperlink in web development.
我正在尝试使单独的文本片段UILabel可点击。我正在寻找的东西通常被称为 Web 开发中的超链接。
<a href="//example.com">Link 1</a>
<a href="//example.com/example">Link 2</a>
<a href="//example.com/other_example">Link 3</a>
Each atag is its own UILabel, and it would ideally open Safari to the specified hrefwhen the text between the tags is clicked.
每个a标签都是它自己的UILabel，理想情况下，href当点击标签之间的文本时，它会打开 Safari 到指定的位置。
I've found a bevy of resources on how to do this sort of thing in Objective-C, but they all seem unnecessarily complicated and don't translate well to Swift (they fit an Objective-C organizational structure that doesn't work well in Swift and goes against the recommended way of using the language).
我找到了很多关于如何在 Objective-C 中做这种事情的资源，但它们似乎都不必要地复杂，并且不能很好地转换为 Swift（它们适合无法很好地工作的 Objective-C 组织结构在 Swift 中，并且违背了使用该语言的推荐方式）。
Here are a few:
以下是一些：
- How to add hyperlink in iPhone app?
- How to make a clickable link inside a NSTextField and Cocoa
- Text as Hyperlink in Objective-C
If I had a 3 UILabels,
如果我有一个3UILabel秒，
Item 1
Item 2
Item 3
第 1 项
第 2 项
第 3 项
then what would be the best "Swift-y" way to make each item open to a different URL in Safari?
那么让每个项目在 Safari 中打开到不同 URL 的最佳“Swift-y”方式是什么？
I could create separate buttons for each, but the UILabels are programmatically populated, so I was thinking that making the text respond to taps might be a better option.
我可以为每个按钮创建单独的按钮，但UILabels 是以编程方式填充的，所以我认为让文本响应点击可能是更好的选择。
回答by Jorge Cordero
Swift 3I created a LinkUILabel class in github:
https://github.com/jorgecsan/LinkUILabelWith this you only need add the url inspectable as the shows the image:
or assign the url variable programmatically:
Swift 3我在 github 中创建了一个 LinkUILabel 类：https:
//github.com/jorgecsan/LinkUILabel有了这个，您只需要添加可检查的 url，如图所示：
或以编程方式分配 url 变量：
linkUILabel.url = "www.example.com"
If you want to implement by your self also I found that solution!:)
如果您想自己实施，我也找到了解决方案！:)
using:
使用：
// This is the label
@IBOutlet weak var label: UILabel!
override func loadView() {
super.loadView()
// This is the key
let tap = UITapGestureRecognizer(target: self, action: #selector(self.onClicLabel(sender:)))
label.isUserInteractionEnabled = true
label.addGestureRecognizer(tap)
}
// And that's the function :)
func onClicLabel(sender:UITapGestureRecognizer) {
openUrl("http://www.google.com")
}
func openUrl(urlString:String!) {
let url = URL(string: urlString)!
if #available(iOS 10.0, *) {
UIApplication.shared.open(url, options: [:], completionHandler: nil)
} else {
UIApplication.shared.openURL(url)
}
}
Hope it helps!:)
希望能帮助到你！：）
回答by Paulw11
The One approach would be something like the following.
The assumptions are:
One 方法类似于以下内容。
假设是：
- self.urls is a string array containing the urls associated with each
UILabel. - Each
UILabeltaghas been set to the corresponding index in the array labelTapped:is set as thetouchUpInsidehandler for the labels.
- self.urls 是一个字符串数组，包含与每个
UILabel. - 每个
UILabeltag都被设置为数组中的对应索引 labelTapped:被设置为touchUpInside标签的处理程序。
import Foundation
import UIKit
class urltest {
var urls:[String]
init() {
self.urls=[String]() // Load URLs into here
}
@IBAction func labelTapped(sender:UILabel!) {
let urlIndex=sender.tag;
if (urlIndex >= 0 && urlIndex < self.urls.count) {
self.openUrl(self.urls[urlIndex]);
}
}
func openUrl(url:String!) {
let targetURL=NSURL.URLWithString(url)
let application=UIApplication.sharedApplication()
application.openURL(targetURL);
}
}
