ios - 如何创建地址簿和联系人

Question

我已经为地址簿和联系人创建了一个代码，用于显示联系人。它工作正常，但突然出现运行时错误。由于我是 ios 新手。我无法找到错误，可以在Stack Overflow 告诉错误。

- (void)viewDidLoad
 {
 [super viewDidLoad];
 self.view = [[UIView alloc] initWithFrame:[[UIScreen mainScreen] applicationFrame]];
 self.view.backgroundColor = [UIColor yellowColor];
 self.navigationItem.leftBarButtonItem = [[[UIBarButtonItem alloc]initWithBarButtonSystemItem:UIBarButtonSystemItemCancel target:self action:@selector(gotohomepage:)]autorelease];
   ABPeoplePickerNavigationController *picker = [[ABPeoplePickerNavigationController alloc] init];
[[picker navigationBar] setBarStyle:UIBarStyleBlack];
picker.peoplePickerDelegate = self;
// Display only a person's phone, email, and birthdate
NSArray *displayedItems = [NSArray arrayWithObjects:[NSNumber numberWithInt:kABPersonPhoneProperty],nil];


picker.displayedProperties = displayedItems;
[self presentModalViewController:picker animated:YES];
[picker release];


}
- (IBAction)gotohomepage:(id)sender
{
[self dismissViewControllerAnimated:YES completion:nil];
}
- (BOOL)peoplePickerNavigationController:(ABPeoplePickerNavigationController *)peoplePicker shouldContinueAfterSelectingPerson:(ABRecordRef)person
{


ABAddressBookRef addressBook = ABAddressBookCreate();

int i;
NSString *strName = @"";
NSString* company = @"";
NSString *address = @"";
NSString *suburb = @"";
NSString *postalcode = @"";
NSString *state = @"";
NSString *country = @"";
NSString *mobile = @"";
NSString *phone = @"";
NSString *emailid = @"";


strName = (NSString *)ABRecordCopyCompositeName((ABRecordRef) person);
CFStringRef name = ABRecordCopyCompositeName((ABRecordRef) person);
company  = (NSString *)ABRecordCopyValue((ABRecordRef) person, kABPersonOrganizationProperty);

NSArray*  allPeople = (NSArray *)ABAddressBookCopyPeopleWithName(addressBook,name);
CFRelease(name);

for (i = 0; i < [allPeople count]; i++)
{
    ABRecordRef record = [allPeople objectAtIndex:i];

    ABMutableMultiValueRef multiValue = ABRecordCopyValue(record, kABPersonAddressProperty);
    for(CFIndex i=0; i<ABMultiValueGetCount(multiValue); i++)
    {
        NSString* HomeLabel = (NSString*)ABMultiValueCopyLabelAtIndex(multiValue, i);
        if([HomeLabel isEqualToString:@"_$!<Home>!$_"])
        {
            CFDictionaryRef dict = ABMultiValueCopyValueAtIndex(multiValue, i);
            address = [NSString stringWithFormat:@"%@", CFDictionaryGetValue(dict, kABPersonAddressStreetKey)];
            suburb = [NSString stringWithFormat:@"%@", CFDictionaryGetValue(dict, kABPersonAddressCityKey)];
            postalcode = [NSString stringWithFormat:@"%@", CFDictionaryGetValue(dict, kABPersonAddressZIPKey)];
            state = [NSString stringWithFormat:@"%@", CFDictionaryGetValue(dict, kABPersonAddressStateKey)];
            country = [NSString stringWithFormat:@"%@", CFDictionaryGetValue(dict, kABPersonAddressCountryKey)];

            CFRelease(dict);
        }
        CFRelease(HomeLabel);
    }
    CFRelease(multiValue);
}
CFRelease(allPeople);


ABMultiValueRef phones =(NSString*)ABRecordCopyValue(person, kABPersonPhoneProperty);
NSString* mobileLabel = nil;
for(CFIndex i = 0; i < ABMultiValueGetCount(phones); i++)
{
    mobileLabel = (NSString*)ABMultiValueCopyLabelAtIndex(phones, i);
    if([mobileLabel isEqualToString:(NSString *)kABPersonPhoneMobileLabel])
    {
        mobile = (NSString*)ABMultiValueCopyValueAtIndex(phones, i);
        NSLog(@"phone   %@",mobile);
    }
    else if ([mobileLabel isEqualToString:(NSString*)kABPersonPhoneIPhoneLabel])
    {
        phone = (NSString*)ABMultiValueCopyValueAtIndex(phones, i);
        NSLog(@"phone   %@",phone);

        CFRelease(mobileLabel);
        break ;
    }
    CFRelease(mobileLabel);

}
CFStringRef value, label;
ABMutableMultiValueRef multi = ABRecordCopyValue(person, kABPersonEmailProperty);
CFIndex count = ABMultiValueGetCount(multi);
if (count == 1)
{
    value = ABMultiValueCopyValueAtIndex(multi, 0);
    emailid = (NSString*) value;
    NSLog(@"self.emailID   %@",emailid);
    CFRelease(value);
}
else
{
    for (CFIndex i = 0; i < count; i++)
    {
        label = ABMultiValueCopyLabelAtIndex(multi, i);
        value = ABMultiValueCopyValueAtIndex(multi, i);

        // check for Work e-mail label
        if (CFStringCompare(label, kABWorkLabel, 0) == 0)
        {
            emailid = (NSString*) value;
            NSLog(@"self.emailID   %@",emailid);
        }
        else if(CFStringCompare(label, kABHomeLabel, 0) == 0)
        {
            emailid = (NSString*) value;
            NSLog(@"self.emailID   %@",emailid);
        }

        CFRelease(label);
        CFRelease(value);
    }
}
CFRelease(multi);    
CFRelease(phones);
CFRelease(addressBook);

[self dismissModalViewControllerAnimated:YES];

return NO;

}

Answer 1

在iOS6上，苹果引入了新的隐私控制，用户可以控制每个应用程序对联系人和日历的访问。所以，在代码端，需要添加一些请求权限的方法。在iOS5或者之前，我们可以随时调用

ABAddressBookRef addressBook = ABAddressBookCreate();

获取地址簿没有任何问题，但是在iOS6中，如果没有权限，这个调用只会返回空指针。这就是为什么我们需要更改获取 ABAddressBookRef 的方法。

此类错误一般出现在没有授予访问通讯录权限的情况下，请添加这段代码

__block BOOL accessGranted = NO;
if (ABAddressBookRequestAccessWithCompletion != NULL) { // we're on iOS 6
    dispatch_semaphore_t sema = dispatch_semaphore_create(0);
    ABAddressBookRequestAccessWithCompletion(addressBook, ^(bool granted, CFErrorRef error) {
        accessGranted = granted;
        dispatch_semaphore_signal(sema);
    });
    dispatch_semaphore_wait(sema, DISPATCH_TIME_FOREVER);
    dispatch_release(sema);   
}
else { // we're on iOS 5 or older
    accessGranted = YES;
}

if (accessGranted) {
    // Do whatever you want here.
}

礼貌:- http://programmerjoe.blogspot.in/2012/10/ios6-permissions-contacts.html

如果这不能解决您的问题，请回复我..