在Swift中使用Fabric - Cannonball的应用程序崩溃？

Question

运行CannonBall时遇到此错误：app here

'[Fabric] Info.plist键的值“Fabric”必须是NSDictionary。'

Appdelegate.swift：

func application(application: UIApplication, didFinishLaunchingWithOptions launchOptions: [NSObject : AnyObject]?) -> Bool {        
    let welcome = "Welcome to Cannonball! Please onboard with the Fabric Mac app. Check the instructions in the README file."
    //assert(NSBundle.mainBundle().objectForInfoDictionaryKey("Fabric") != nil, welcome)

    Fabric.with([Crashlytics(), Twitter(), Digits(), MoPub()])//error here


    if Twitter.sharedInstance().session() == nil && Digits.sharedInstance().session() == nil {
        let storyboard = UIStoryboard(name: "Main", bundle: nil)
        let signInViewController: AnyObject! = storyboard.instantiateViewControllerWithIdentifier("SignInViewController")
        window?.rootViewController = signInViewController as? UIViewController
    }

    return true
}

我得到的建议是“缺少来自info.plist的Fabric APIKey”。 但我在info.plist中找不到Fabric键。

的Info.plist：

Answer 1

将此代码添加到.plist文件中：

<key>Fabric</key>
<dict>
    <key>APIKey</key>
    <string>Your-key</string>
    <key>Kits</key>
    <array>
        <dict>
            <key>KitInfo</key>
            <dict/>
            <key>KitName</key>
            <string>Crashlytics</string>
        </dict>
        <dict>
            <key>KitInfo</key>
            <dict>
                <key>consumerKey</key>
                <string>Your-consumerKey</string>
                <key>consumerSecret</key>
                <string>Your-consumerSecret</string>
            </dict>
            <key>KitName</key>
            <string>Twitter</string>
        </dict>
        <dict>
            <key>KitInfo</key>
            <dict>
                <key>consumerKey</key>
                <string>Your-consumerKey</string>
                <key>consumerSecret</key>
                <string>Your-consumerSecret</string>
            </dict>
            <key>KitName</key>
            <string>Digits</string>
        </dict>
    </array>
</dict>