Question

我有两个测试应用程序：App1＆amp; APP2。

App1从文本字段中获取String并触发方法：

@IBAction func openApp(sender: AnyObject) { 
    let url1 = ("app2://com.application.started?displayText="+textToSend.text!)
    let url2 = url1.stringByAddingPercentEncodingWithAllowedCharacters(NSCharacterSet.URLQueryAllowedCharacterSet())
    UIApplication.sharedApplication().openURL(NSURL(string: url2!)!)
}

实际打开的App2只有标签，应该更改为通过url发送的文本，代码在AppDelegate.swift中：

func application(app: UIApplication, openURL url: NSURL, options: [String : AnyObject]) -> Bool {
    let url = url.standardizedURL
    let query = url?.query
    ViewController().labelToDisplayResult.text = query
    return true;
}

不幸的是，我试图将URL的结果传递给实际标签的行给了我这个错误：

EXC_BAD_INSTRUCTION (CODE=EXC_I386_INVOP SUBCODE=0x0)

但是我确实拥有App2中的所有数据，因为我可以在调试器中看到它们的值：

url NSURL   "app2://com.application.started?displayText=564315712437124375" 0x00007fa4e3426320
query   String? "displayText=564315712437124375"

我知道为什么会收到此错误？

...谢谢

Answer 1

您的错误

ViewController().labelToDisplayResult.text = query

ViewController()创建ViewController的新实例，而不是从storyboard加载的实例。我猜labelToDisplayResult是一个插座，所以它是零，所以你得到EXC_BAD_INSTRUCTION (CODE=EXC_I386_INVOP SUBCODE=0x0)

这是我通常做的处理openURL方案，需要考虑两种状态：

目标应用程序之前已启动，因此当打开网址时，目标应用程序处于后台或非活动状态
目标应用未启动，因此当打开网址时，目标应用根本没有运行

在Appdelegate中

class AppDelegate: UIResponder, UIApplicationDelegate {
var openUrl:NSURL? //This is used when to save state when App is not running before the url trigered
var window: UIWindow?


func application(app: UIApplication, openURL url: NSURL, options: [String : AnyObject]) -> Bool {
    let url = url.standardizedURL
    NSNotificationCenter.defaultCenter().postNotificationName("HANDLEOPENURL", object:url!)
    self.openUrl = url
    return true;
}

func application(application: UIApplication, didFinishLaunchingWithOptions launchOptions: [NSObject: AnyObject]?) -> Bool {

    return true
}
}

然后在ViewController句柄中打开openURL

class ViewController: UIViewController {

@IBOutlet weak var testLabel: UILabel!
override func viewDidLoad() {
    super.viewDidLoad()
     NSNotificationCenter.defaultCenter().addObserver(self, selector: "handleOpenURL:", name:"HANDLEOPENURL", object: nil)
    let delegate = UIApplication.sharedApplication().delegate as? AppDelegate
    if let url = delegate?.openUrl{
       testLabel.text = url.description 
        delegate?.openUrl = nil 
    }
}
func handleOpenURL(notification:NSNotification){
    if let url = notification.object as? NSURL{
        testLabel.text = url.description
    }
}
deinit{
    NSNotificationCenter.defaultCenter().removeObserver(self, name: "HANDLEOPENURL", object:nil)
}

}

使用URL方案

1 个答案: