Question

我看到一些关于这个问题的帖子，但仍然没有得到它。

我为包含应用程序设置了URLidentifier - KbrdApp。如果我的包含应用程序启动，则返回标记“1”，表示“准备好”。

这是来自viewDidLoad的自定义键盘扩展程序的代码：

UIWebView * webView = [[UIWebView alloc] initWithFrame:CGRectMake(0, 0, 100, 100)];
NSString *customURL = @"KbrdApp://";
NSURLRequest *request = [[NSURLRequest alloc] initWithURL:[NSURL URLWithString:customURL]];
[webView loadRequest:request];

我已经读过NSExtensionContext仅适用于今日扩展程序，因此我尝试了UIWebView。但它不起作用。

有什么问题？

Answer 1

问题中提到的方法在iOS 8.0中停止工作。

目前（iOS 9.2），您可以使用多种方式访问共享的UIApplication实例及其隐藏的openURL方法。

就个人而言，我更喜欢我开发的这种方法：

// Usage:
// UIApplication.sharedApplication().openURL(NSURL(string: "your-app-scheme://")!)

extension UIApplication {

    public static func sharedApplication() -> UIApplication {
        guard UIApplication.respondsToSelector("sharedApplication") else {
            fatalError("UIApplication.sharedKeyboardApplication(): `UIApplication` does not respond to selector `sharedApplication`.")
        }

        guard let unmanagedSharedApplication = UIApplication.performSelector("sharedApplication") else {
            fatalError("UIApplication.sharedKeyboardApplication(): `UIApplication.sharedApplication()` returned `nil`.")
        }

        guard let sharedApplication = unmanagedSharedApplication.takeUnretainedValue() as? UIApplication else {
            fatalError("UIApplication.sharedKeyboardApplication(): `UIApplication.sharedApplication()` returned not `UIApplication` instance.")
        }

        return sharedApplication
    }

    public func openURL(url: NSURL) -> Bool {
        return self.performSelector("openURL:", withObject: url) != nil
    }

}

Answer 2

这是我的obj c语言版本。适用于IOS 11.4.1，XCode 9.4.1

UIApplication类的类别：

// UIApplication+UpenURL.h file
@interface UIApplication (OpenUrlHelper)

+ (UIApplication *)sharedApp;
- (void)openUrl:(NSURL *)url withOptions:(NSDictionary *)options andComplition:(void(^)(BOOL))complition;

@end

// UIApplication+UpenURL.m file
#import "UIApplication+UpenURL.h"
@implementation UIApplication (OpenUrlHelper)

+ (UIApplication *)sharedApp {
    __weak UIApplication *sharedApp = [UIApplication performSelector:@selector(sharedApplication)];
    return sharedApp;
}

- (void)openUrl:(NSURL *)url withOptions:(NSDictionary *)options andComplition:(void(^)(BOOL))complition {
    if (@available (iOS 10, *)) {
        [self performSelector:@selector(openURL:options:completionHandler:)
                   withObject:url
                   withObject:options
                   withObject:complition];
    } else {
        [self performSelector:@selector(openURL:) withObject:url];
    }
}

- (id)performSelector:(SEL)selector withObject:(id)obj1 withObject:(id)obj2 withObject:(id)obj3 {
    NSMethodSignature *signature = [self methodSignatureForSelector:selector];
    if (!signature)
        return nil;

    NSInvocation* invocation = [NSInvocation invocationWithMethodSignature:signature];
    [invocation setTarget:self];
    [invocation setSelector:selector];
    [invocation setArgument:&obj1 atIndex:2];
    [invocation setArgument:&obj2 atIndex:3];
    [invocation setArgument:&obj3 atIndex:4];
    [invocation invoke];
    if (signature.methodReturnLength) {
        id anObject;
        [invocation getReturnValue:&anObject];
        return anObject;
    }
    return nil;
}

@end

然后将其保存在需要的文件中

#import "UIApplication+UpenURL.h"

并使用如下代码：

NSURL *url = [NSURL URLWithString:@"yourapp://"];
[[UIApplication sharedApp] openUrl:url withOptions:@{} andComplition:nil];

如何从自定义键盘扩展启动包含应用程序？

2 个答案: