在iOS 6中，以前工作正常的訪問通訊錄的iPhone程序可能會出錯，現(xiàn)象是程序啟動時不提醒用戶是否允許程序訪問通訊錄，同時在“設置->隱私->通訊錄”中看不到你的程序。另外，對通訊錄進行操作的代碼會報類似于以下消息的錯誤：

?

?

    Could not compile statement for query (ABCCopyArrayOfAllInstancesOfClassInSourceMatchingProperties):

SELECT ROWID, Name, ExternalIdentifier, Type, ConstraintsPath, ExternalModificationTag, ExternalSyncTag, AccountID, Enabled, SyncData, MeIdentifier, Capabilities FROM ABStore WHERE Enabled = ?;
  

其原因是iOS 6加強了通訊錄訪問控制，要求開發(fā)人員顯式聲明需要訪問通訊錄，方法是調用 ? ABAddressBookRequestAccessWithCompletion

?

方法，具體參見官方文檔：

http://developer.apple.com/library/ios/#releasenotes/General/RN-iOSSDK-6_0/index.html

?

下面是對應的樣例代碼，一般來講需要將這段代碼放置在程序啟動部分，在程序啟動過程中提示用戶本程序需要訪問設備上的通訊錄：

?

    ABAddressBookRef addressBook = ABAddressBookCreate();

    

        

        __block BOOL accessGranted = NO;

    if (ABAddressBookRequestAccessWithCompletion != NULL) {

        

        // we're on iOS 6

        NSLog(@"on iOS 6 or later, trying to grant access permission");

        

        dispatch_semaphore_t sema = dispatch_semaphore_create(0);

        ABAddressBookRequestAccessWithCompletion(addressBook, ^(bool granted, CFErrorRef error) {

            accessGranted = granted;

            dispatch_semaphore_signal(sema);

        });

        dispatch_semaphore_wait(sema, DISPATCH_TIME_FOREVER);

        dispatch_release(sema);

    }

    else { // we're on iOS 5 or older

        

        NSLog(@"on iOS 5 or older, it is OK");

        accessGranted = YES;

    }

    

    if (accessGranted) {

        

        NSLog(@"we got the access right");

    }

    


  

?

?

iOS6 中如何獲得通訊錄訪問權限