找到了答案:iOS 开发人员库中有很好的记录:http:
//developer.apple.com/library/ios/#samplecode/QuickContacts/Listings/Classes_QuickContactsViewController_m.html#//apple_ref/doc/uid/DTS40009475-Classes_QuickContactsViewController_m-DontLinkElementID_6
这是我实现以将 ABPersonRecordRef 作为对象返回的示例代码。我遇到的错误与返回 ABPersonRecordRef 对象后没有保留它有关。
- (id)personRecordUsingModelObj:(id)modelObj {
ABRecordRef aContact = ABPersonCreate();
CFErrorRef anError = NULL;
NSString *name = [NSString stringWithFormat:@"%@", [modelObj name]];
ABRecordSetValue(aContact, kABPersonOrganizationProperty, name, &anError);
ABMultiValueRef phone = ABMultiValueCreateMutable(kABMultiStringPropertyType);
ABMultiValueAddValueAndLabel(phone, [modelObj phone], kABPersonPhoneMainLabel, NULL);
ABRecordSetValue(aContact, kABPersonPhoneProperty, phone, &anError);
CFRelease(phone);
NSString *address = [NSString stringWithFormat:@"%@ %@", [modelObj addr1], [modelObj addr2]];
NSMutableDictionary *dictionaryAddress = [[NSMutableDictionary alloc] initWithCapacity:0];
[dictionaryAddress setObject:address forKey:(NSString *)kABPersonAddressStreetKey];
[dictionaryAddress setObject:@"us" forKey:(NSString *)kABPersonAddressCountryCodeKey];
ABMutableMultiValueRef address = ABMultiValueCreateMutable(kABDictionaryPropertyType);
ABMultiValueAddValueAndLabel(address, dictionaryAddress, kABPersonAddressStreetKey, NULL);
[dictionaryAddress release];
ABRecordSetValue(aContact, kABPersonAddressProperty, address, &anError);
CFRelease(address);
if (anError) {
aContact = nil;
}
[(id)aContact autorelease];
return (id)aContact;
}