iOS10 中的 OpenURL

Question

显然 OpenURL 已在 iOS 中弃用 10. 有没有人有任何文档说明原因或可以解释下一步该做什么？我已经查看了 Apple 网站并发现了一些与 OpenURL 有关的内容，这就是他们现在说要使用的内容：

UIApplication.shared().open(url: URL, options: [String: AnyObject], completionHandler: ((Bool) -> Void)?)

有没有人有任何证据表明这是在 Swift 3.0 中使用 OpenURL 的新方法？此外，options: 和 completionHandler: 参数分别使用什么值？

Answer 1

The new UIApplication method openURL:options:completionHandler:, which is executed asynchronously and calls the specified completion handler on the main queue (this method replaces openURL:).

这是在其他框架更改 > UIKit 下，位于：https://developer.apple.com/library/prerelease/content/releasenotes/General/WhatsNewIniOS/Articles/iOS10.html

Answer 2

空选项字典将导致与 openUrl.

相同的行为

否则：

+-------------------------------------------------+----------------------------------------------------------------------------------------------------------------------------------------------+
| UIApplicationOpenURLOptionsSourceApplicationKey | NSString containing the bundle ID of the originating application                                                                             |
+-------------------------------------------------+----------------------------------------------------------------------------------------------------------------------------------------------+
| UIApplicationOpenURLOptionsAnnotationKey        | property-list typed object corresponding to what the originating application passed in UIDocumentInteractionController's annotation property |
+-------------------------------------------------+----------------------------------------------------------------------------------------------------------------------------------------------+
| UIApplicationOpenURLOptionsOpenInPlaceKey       | bool NSNumber, set to YES if the file needs to be copied before use                                                                          |
+-------------------------------------------------+----------------------------------------------------------------------------------------------------------------------------------------------+

来自UIApplication.h

// Options are specified in the section below for openURL options. An empty options dictionary will result in the same
// behavior as the older openURL call, aside from the fact that this is asynchronous and calls the completion handler rather
// than returning a result.
// The completion handler is called on the main queue.
- (void)openURL:(NSURL*)url options:(NSDictionary<NSString *, id> *)options completionHandler:(void (^ __nullable)(BOOL success))completion NS_AVAILABLE_IOS(10_0) NS_EXTENSION_UNAVAILABLE_IOS("");

UIKIT_EXTERN UIApplicationOpenURLOptionsKey const UIApplicationOpenURLOptionsSourceApplicationKey NS_SWIFT_NAME(sourceApplication) NS_AVAILABLE_IOS(9_0);   // value is an NSString containing the bundle ID of the originating application
UIKIT_EXTERN UIApplicationOpenURLOptionsKey const UIApplicationOpenURLOptionsAnnotationKey NS_SWIFT_NAME(annotation) NS_AVAILABLE_IOS(9_0);   // value is a property-list typed object corresponding to what the originating application passed in UIDocumentInteractionController's annotation property
UIKIT_EXTERN UIApplicationOpenURLOptionsKey const UIApplicationOpenURLOptionsOpenInPlaceKey NS_SWIFT_NAME(openInPlace) NS_AVAILABLE_IOS(9_0);   // value is a bool NSNumber, set to YES if the file needs to be copied before use

Answer 3

Swift 3+:

func open(scheme: String) {
   if let url = URL(string: scheme) {
      if #available(iOS 10, *) {
         UIApplication.shared.open(url, options: [:],
           completionHandler: {
               (success) in
                  print("Open \(scheme): \(success)")
           })
     } else {
         let success = UIApplication.shared.openURL(url)
         print("Open \(scheme): \(success)")
     }
   }
 }

用法：

open(scheme: "tweetbot://timeline")

Source

Answer 4

如果您的应用仍然支持 iOS 9 或更低版本，请继续使用旧的 openURL。如果您的部署目标是 iOS 10.

，您应该只移动到新的。

Answer 5

快速修复：

// Objective-C
UIApplication *application = [UIApplication sharedApplication];
[application openURL:URL options:@{} completionHandler:nil];

// Swift
UIApplication.shared.open(url, options: [:], completionHandler: nil)

完整答案：

http://useyourloaf.com/blog/openurl-deprecated-in-ios10/

致谢：基思·哈里森 (useyourloaf.com)

Answer 6

让实际：

[String: AnyObject] = ["xxx key": "xxx value" as AnyObject, "yyy key": "yyy value" as AnyObject]

UIApplication.shared.open(URL(string: "http:google.com")!, options: actual, completionHandler: {(true) -> Swift.Void in
print("Refresh")
})

其中 xxx 和 yyy 是您要打印的任何字符串或将它们留空。

Answer 7

在加载 url 之前，您需要先进行一些检查。请检查以下代码。

if ([[UIApplication sharedApplication] canOpenURL:[NSURL URLWithString:@"https://www.gmail.com"]]){
    [[UIApplication sharedApplication] openURL:[NSURL URLWithString:@"https://www.gmail.com"] options:@{} completionHandler:^(BOOL success) {
                                //completion codes here
                            }];
}

希望对您有所帮助。

Answer 8

您可以使用函数打开设置：

func showAlert(title:String, message:String){
        let alert = UIAlertController(title: title,
                                      message: message,
                                      preferredStyle: UIAlertController.Style.alert)

        let okAction = UIAlertAction(title: "OK", style: .cancel, handler: nil)
        alert.addAction(okAction)

        let settingsAction = UIAlertAction(title: "Settings", style: .default, handler: { _ in
            // Take the user to Settings app to possibly change permission.
            guard let settingsUrl = URL(string: UIApplication.openSettingsURLString) else { return }
            if UIApplication.shared.canOpenURL(settingsUrl) {
                if #available(iOS 10.0, *) {
                    UIApplication.shared.open(settingsUrl, completionHandler: { (success) in
                        // Finished opening URL
                    })
                } else {
                    // Fallback on earlier versions
                    UIApplication.shared.openURL(settingsUrl)
                }
            }
        })
        alert.addAction(settingsAction)
        present(alert, animated: true, completion: nil)
    }

像下面给出的那样调用这个函数

showAlert(title: "Unable to access the Photos", message: "To enable access, go to Settings > Privacy > Photos and turn on Photos access for this app.")

iOS10 中的 OpenURL

OpenURL in iOS10

xcode

openurl

ios

swift