I'm working on my app, I wanted to add AdMob banner like I always do without any problem before, but now on this new app it doesn't show up.
The banner ad is in one of the UITableViewCell's, and I get this log message:
-canOpenURL: failed for URL: "kindle://home" - error: "(null)"
I have disabled Bitcode and I have added the App Transport Security and NSAllowsArbitraryLoads in the info.plist file.
Update to the latest AdMob SDK. This issue was fixed in 7.5.0:
SDK no longer uses UIApplication's canOpenURL method on iOS 9 devices.
AdMob for iOS Release Notes
I simply pasted this into my info.plist file and it all worked
<key>NSAppTransportSecurity</key>
<dict>
<key>NSAllowsArbitraryLoads</key>
<true/>
</dict>
as per https://developers.google.com/admob/ios/ios9
Related
I have an app with AdMob. In plist I added apps ID and in code added banner ID.
<key>GADApplicationIdentifier</key>
<string>ca-app-pub-XXXX~XXXX</string>
<key>SKAdNetworkItems</key>
<array>
<dict>
<key>SKAdNetworkIdentifier</key>
<string>cstr6suwn9.skadnetwork</string>
</dict>
</array>
bannerView.adUnitID = "YYYY/YYYY"
Test ads I can see, if I explicitly add testDeviceIdentifiers, otherwise I am getting an error To get test ads on this device, set: GADMobileAds.sharedInstance.requestConfiguration.testDeviceIdentifiers.
The App was published on the App Store few days ago. I was thinking AdMob can determine it and show the real Ads, but it did not happen. What am I missing?
Well the ads appeared after few days. Really weird that it took so much time.
I'm using UIWebView XCode Version 8.3. My console continues to show me this error when I start the app and I do not understand what it means. In my UIViewController there is no code so I do not understand why XCode continues to show me this error.
I have only one place in my webview viewcontroller by Storyboard.
This is the error:-
2017-04-07 23:54:50.081592+0200 Conquers[1647:697735]
libMobileGestalt MobileGestaltSupport.m:153: pid 1647 (Conquers) does
not have sandbox access for frZQaeyWLUvLjeuEK43hmg and
IS NOT appropriately entitled
2017-04-07 23:54:50.081651+0200 Conquers[1647:697735] libMobileGestalt
MobileGestalt.c:550: no access to InverseDeviceID (see
)
Just by searching on the web when I came across this issue and found that can be an Apple error (no bugs or crash are appearing with this error in my app).
If the error occurs only when you set up the UIWebView on your app. I would check if the app has configured correctly the "App transport security Settings" in your .plist
if you inspect your .plist you can add it by hand like:
<key>NSAppTransportSecurity</key>
<dict>
<key>NSAllowsArbitraryLoads</key>
<true/>
<key>NSExceptionDomains</key>
<dict>
<key>add-your-site-here-if-needed</key>
<dict>
<key>NSExceptionAllowsInsecureHTTPLoads</key>
<true/>
<key>NSIncludesSubdomains</key>
<true/>
</dict>
</dict>
</dict>
This happens to me also when I added some third-party code like FirebaseSDK. And it suddenly stops appearing after I setup correctly my URL schemes for all my environments. So I´m assuming that this error appears when something is not fully/correctly implemented like in a snowball effect by Apple side code.
NOTE: You can also try to go back a few steps back on when the error occurs to narrow down the possibilities of occurrences of this error.
I have develop one application in which there are so many photos and videos.
I have develop this app in xcode6.3 and it will work perfectly but now i have installed new xcode7.1 and when i'm trying to run this app in xcode7.1 it will run perfectly but i haven't seen any videos and images of my app.
Is that any chnage that i have to do in my code or is it any other problem
Starting from Xcode7 Apple mandates that all url access should be HTTPS as a part of Apple Transport Security (ATS)
To overcome this you have to add the following item in your app info.plist file
You can bypass this by adding this key to your info.plist of the project
<key>NSAppTransportSecurity</key>
<dict>
<key>NSAllowsArbitraryLoads</key>
<true/>
</dict>
for more details about ATS refer NSURLSession/NSURLConnection HTTP load failed on iOS 9
http://ste.vn/2015/06/10/configuring-app-transport-security-ios-9-osx-10-11/
Following code was working fine with iOS 8.4 and earlier, but from iOS 9.0 it starts throwing warning as well giving false always.
[[UIApplication sharedApplication] canOpenURL:[NSURL URLWithString:#"comgooglemaps-x-callback://"]]
It is throwing warning as:
-canOpenURL: failed for URL: "comgooglemaps-x-callback://" - error: "This app is not allowed to query for scheme comgooglemaps-x-callback"
Do anyone having any idea about, why it is failing in iOS 9 and why throwing error as This app is not allowed to query for scheme comgooglemaps-x-callback, is there any permissions required to query it?
EDIT
I have added following items in my info.plist, but it is still having same problem.
<key>LSApplicationQueriesSchemes</key>
<array>
<string>comgooglemaps</string>
<string>comgooglemaps-x-callback</string>
</array>
You have to add a LSApplicationQueriesSchemes entry to your Info.plist and list the schemes that you want your app to query.
<key>LSApplicationQueriesSchemes</key>
<array>
<string>urlscheme</string>
<string>urlscheme2</string>
<string>urlscheme3</string>
<string>urlscheme4</string>
</array>
see http://useyourloaf.com/blog/querying-url-schemes-with-canopenurl.html (also contains a Link to the corresponding WWDC15-Session-Video)
I have been trying to deal with this problem for a while now, and I'm hoping someone has a solution that I haven't yet seen.
I am using XCode 7 and am testing my app on an iPad (ios 9 beta). I have tried adding the code below to every instance of Info.plist and it doesn't work.
<key>NSAppTransportSecurity</key>
<dict>
<key>NSAllowsArbitraryLoads</key>
<true/>
</dict>
I have tried decreasing the Deployment Target to 8.4. I have tried restarting both my laptop I am developing on and the iPad.
I just don't know what else to do. Thanks.
You may just be adding it in an invalid position, it should be inside the main <dict> tag. Try adding as the first element in the first dictionary like this.