将通讯录联系人存储到nsdictionary中 [英] storing addressbook contacts into a nsdictionary

问题描述

我仍在尝试使用NSDictionaries来解决问题，并且遇到了我认为需要使用NSDictionaries的情况。本质上，我想将与每个联系人关联的所有电话号码存储在字典中。到目前为止，我有这个：

I'm still trying to wrap my head around using NSDictionaries, and have come into a situation where I believe I need to use one. essentially, I would like to store all the phone numbers associated with each contact into a dictionary. so far I have this:

ABAddressBookRef addressBook = ABAddressBookCreate();
NSArray *thePeople = (NSArray *)ABAddressBookCopyArrayOfAllPeople(addressBook); 
for (id person in thePeople)
{
    ABMultiValueRef phones =(NSString*)ABRecordCopyValue(person, kABPersonPhoneProperty);
    NSString* name = (NSString *)ABRecordCopyCompositeName(person); 
    for (CFIndex i = 0; i < ABMultiValueGetCount(phones); i++)
    {
        NSString *phone = [(NSString *)ABMultiValueCopyValueAtIndex(phones,i) autorelease];
    }
}

我想知道如何使用nsdictionary来存储每个人员，然后是与该人员相关的每个电话值的数组。

I was wondering how to use a nsdictionary to store each person, and then an array of each phone value that's associated with that person.

推荐答案

您要做什么？

您可以将所有姓名和电话号码放入这样的plist中：

You can put all names and phonenumbers into a plist like this:

ABAddressBookRef addressBook = ABAddressBookCreate();
NSArray *thePeople = (NSArray *)ABAddressBookCopyArrayOfAllPeople(addressBook);
NSMutableArray* allPeoplesDicts = [NSMutableArray array];
for (id person in thePeople)
{
    ABMultiValueRef phones =(NSString*)ABRecordCopyValue(person, kABPersonPhoneProperty);
    NSString* name = (NSString *)ABRecordCopyCompositeName(person);
    NSMutableArray* phones = [[NSMutableArray alloc] init];
    for (CFIndex i = 0; i < ABMultiValueGetCount(phones); i++)
    {
        NSString *phone = [(NSString *)ABMultiValueCopyValueAtIndex(phones,i) autorelease];
        [phones addObject:phone];
    }
    NSDictionary* personDict = [[NSDictionary alloc] initWithObjectsAndKeys:name,@"Name",phones,@"PhoneNumbers",nil];
    [phones release];
    [allPeoplesDicts addObject:personDict];
    [personDict release];
}

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