我为地址簿和联系人创建了一个显示联系人的代码。它运行正常但突然出现运行时错误。因为我是ios的新手。我无法找到任何一个错误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;
}
答案 0 :(得分:1)
在iOS6上,Apple引入了新的隐私控制,用户可以通过每个应用程序控制联系人和日历的访问权限。因此,在代码方面,您需要添加一些方法来请求权限。在iOS5或之前,我们总是可以调用
ABAddressBookRef addressBook = ABAddressBookCreate();
获取地址簿没有任何问题,但在iOS6中,如果您没有权限,此调用将只返回空指针。这就是为什么我们需要更改方法来获取ABAddressBookRef。
当未授予访问地址簿权限时,通常会出现此类错误。在这种情况下,请添加此段代码
__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...
答案 1 :(得分:0)
错误在于,因为在IOS6上,Apple添加了一些功能,没有任何人有权在没有所有者许可的情况下获取内容。