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在我的應用程序中,如果我點擊一個按鈕,我需要顯示所有手機接觸像whatsapp ...如何實現AddressBook框架與此?或者,我們可以使用任何其他框架中顯示所有設備聯繫人...在ios中使用系統框架顯示iPhone聯繫人
在我的應用程序中,如果我點擊一個按鈕,我需要顯示所有手機接觸像whatsapp ...如何實現AddressBook框架與此?或者,我們可以使用任何其他框架中顯示所有設備聯繫人...在ios中使用系統框架顯示iPhone聯繫人
我們可以獲取使用接觸框架的接觸:步幅>
添加兩個文件在您的.h文件中:
添加流動代碼
-(void)loadContactList
{
@try {
CNAuthorizationStatus status = [CNContactStore authorizationStatusForEntityType:CNEntityTypeContacts];
if(status == CNAuthorizationStatusDenied || status == CNAuthorizationStatusRestricted)
{
NSLog(@"access denied");
}
else
{
//Create repository objects contacts
CNContactStore *contactStore = [[CNContactStore alloc] init];
//Select the contact you want to import the key attribute (https://developer.apple.com/library/watchos/documentation/Contacts/Reference/CNContact_Class/index.html#//apple_ref/doc/constant_group/Metadata_Keys)
NSArray *keys = [[NSArray alloc]initWithObjects:CNContactIdentifierKey, CNContactEmailAddressesKey, CNContactBirthdayKey, CNContactImageDataKey, CNContactPhoneNumbersKey,CNContactViewController.descriptorForRequiredKeys,nil];
// Create a request object
CNContactFetchRequest *request = [[CNContactFetchRequest alloc] initWithKeysToFetch:keys];
request.predicate = nil;
[contactStore enumerateContactsWithFetchRequest:request
error:nil
usingBlock:^(CNContact* __nonnull contact, BOOL* __nonnull stop)
{
// Contact one each function block is executed whenever you get
NSString *phoneNumber = @"";
if(contact.phoneNumbers)
phoneNumber = [[[contact.phoneNumbers firstObject] value] stringValue];
NSLog(@"phoneNumber = %@", phoneNumber);
NSLog(@"givenName = %@", contact.givenName);
NSLog(@"familyName = %@", contact.familyName);
NSLog(@"email = %@", contact.emailAddresses);
[contactList addObject:contact];
}];
}
} @catch (NSException *exception) {
NSLog(@"Exception:%@",exception.reason);
}
}
調用此方法鑑於沒有負載或沒有視圖出現。
[網上有很多例子..](https://www.google.com.kw/search?rlz=1C5CHFA_enKW556KW556&espv=2&q=import+contacts+programmatically+ios&oq=import+contacts+programmatically+ios&gs_l= serp.3 ... 8379.16207.0.16443.9.9.0.0.0.0.208.795.0j2j2.4.0 .... 0 ... 1c.1.64.serp..5.3.585 ... 0i22i30k1j0i7i30k1j0i7i5i30k1.1GZZr9YVwOo)請在您之前搜索張貼在這裏.. –
是的,它的作品!謝謝 – Sivagami
@SivagamiSundari如果我的回答對你有幫助,那麼請給它正確的標記,這樣可以幫助其他用戶。 –