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在我的应用程序中,如果我点击一个按钮,我需要显示所有手机接触像whatsapp ...如何实现AddressBook框架与此?或者,我们可以使用任何其他框架中显示所有设备联系人...在ios中使用系统框架显示iPhone联系人
在我的应用程序中,如果我点击一个按钮,我需要显示所有手机接触像whatsapp ...如何实现AddressBook框架与此?或者,我们可以使用任何其他框架中显示所有设备联系人...在ios中使用系统框架显示iPhone联系人
我们可以获取使用接触框架的接触:步幅>
添加两个文件在您的.h文件中:
添加流动代码
-(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);
}
}
调用此方法鉴于没有负载或没有视图出现。
[网上有很多例子..](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)请在您之前搜索张贴在这里.. –
是的,它的作品!谢谢 – Sivagami
@SivagamiSundari如果我的回答对你有帮助,那么请给它正确的标记,这样可以帮助其他用户。 –