is there any best way to listout Property of Contact in detail?
like it is working in IOS 6.1 and earlier Version.
//
// ABPersonViewController.h
// AddressBookUI
//
// Copyright (c) 2010 Apple Inc. All rights reserved.
//
#import <UIKit/UIViewController.h>
#import <AddressBook/AddressBook.h>
@interface ABPersonViewController : UIViewController <UIViewControllerRestoration>
// ABPersonViewController does not support subclassing in iOS 7.0 and later. A nil instance will be returned.
答案 0 :(得分:0)
我找到了解决方案:请参阅source
有一些已弃用的功能,但我们可以解决它, 通过在“ABContactsHelper”类中替换下面的函数
+ (ABAddressBookRef) addressBook
{
#ifdef __IPHONE_6_0
return ABAddressBookCreateWithOptions(NULL, NULL);
#else
return ABAddressBookCreate();
#endif
}
在调用“ABAddressBookCreate()”的地方使用它。 喜欢
ABAddressBookRef addressBook = [ABContactsHelper addressBook];