我正在开发一个应用程序,我想从iphone中的联系人列表中获取正在使用相同应用程序的联系人。
怎么做?任何示例代码或链接?
请帮帮我。
注意:我不想在我的ios应用程序中获取所有联系人,我 只想获取使用相同应用程序的联系人。
答案 0 :(得分:0)
首先导入AddressBook框架
然后调用这两个函数
-(void)AddressBookFetch{
ABAuthorizationStatus status = ABAddressBookGetAuthorizationStatus();
if (status == kABAuthorizationStatusDenied || status == kABAuthorizationStatusRestricted) {
// if you got here, user had previously denied/revoked permission for your
// app to access the contacts, and all you can do is handle this gracefully,
// perhaps telling the user that they have to go to settings to grant access
// to contacts
[[[UIAlertView alloc] initWithTitle:nil message:@"This app requires access to your contacts to function properly. Please visit to the \"Privacy\" section in the iPhone Settings app." delegate:nil cancelButtonTitle:@"OK" otherButtonTitles:nil] show];
return;
}
CFErrorRef error = NULL;
ABAddressBookRef addressBook = ABAddressBookCreateWithOptions(NULL, &error);
if (!addressBook) {
NSLog(@"ABAddressBookCreateWithOptions error: %@", CFBridgingRelease(error));
return;
}
ABAddressBookRequestAccessWithCompletion(addressBook, ^(bool granted, CFErrorRef error) {
if (error) {
NSLog(@"ABAddressBookRequestAccessWithCompletion error: %@", CFBridgingRelease(error));
}
if (granted) {
// if they gave you permission, then just carry on
[self listPeopleInAddressBook:addressBook];
} else {
// however, if they didn't give you permission, handle it gracefully, for example...
dispatch_async(dispatch_get_main_queue(), ^{
// BTW, this is not on the main thread, so dispatch UI updates back to the main queue
[[[UIAlertView alloc] initWithTitle:nil message:@"This app requires access to your contacts to function properly. Please visit to the \"Privacy\" section in the iPhone Settings app." delegate:nil cancelButtonTitle:@"OK" otherButtonTitles:nil] show];
});
}
CFRelease(addressBook);
});
}
然后按此功能列出联系人
- (void)listPeopleInAddressBook:(ABAddressBookRef)addressBook
{
NSArray *allPeople = CFBridgingRelease(ABAddressBookCopyArrayOfAllPeople(addressBook));
numberOfPeople = [allPeople count];
for (NSInteger i = 0; i < numberOfPeople; i++) {
ABRecordRef person = (__bridge ABRecordRef)allPeople[i];
NSString *firstName = CFBridgingRelease(ABRecordCopyValue(person, kABPersonFirstNameProperty));
NSString *lastName = CFBridgingRelease(ABRecordCopyValue(person, kABPersonLastNameProperty));
NSLog(@"Name:%@ %@", firstName, lastName);
if (firstName==nil) {
firstName=@"";
}
[nm addObject:firstName];
if (lastName==nil) {
lastName=@"";
}
[ttl addObject:lastName];
ABMultiValueRef phoneNumbers = ABRecordCopyValue(person, kABPersonPhoneProperty);
// CFIndex numberOfPhoneNumbers = ABMultiValueGetCount(phoneNumbers);
//for (CFIndex i = 0; i < numberOfPhoneNumbers; i++) {
NSString *phoneNumber = CFBridgingRelease(ABMultiValueCopyValueAtIndex(phoneNumbers, 0));
NSLog(@" phone:%@", phoneNumber);
if (phoneNumber==nil) {
phoneNumber=@"";
}
[phn addObject:phoneNumber];
// }
// CFRelease(phoneNumbers);
NSLog(@"=============================================");
}
}
答案 1 :(得分:0)
没有后端是不可能的,你不能只在ios应用程序中做到。