在iOS 6中,以前工作正常的访问通讯录的iPhone程序可能会出错,现象是程序启动时不提醒用户是否允许程序访问通讯录,同时在“设置->隐私->通讯录”中看不到你的程序。另外,对通讯录进行操作的代码会报类似于以下消息的错误:
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加强了通讯录访问控制,要求开发人员显式声明需要访问通讯录,方法是调用
ABAddressBookRequestAccessWithCompletion
方法,具体参见官方文档:
http://developer.apple.com/library/ios/#releasenotes/General/RN-iOSSDK-6_0/index.html
下面是对应的样例代码,一般来讲需要将这段代码放置在程序启动部分,在程序启动过程中提示用户本程序需要访问设备上的通讯录:
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"); }
作者:keyboardOTA 发表于2013-3-24 0:39:06 原文链接
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