in IOS,how to remove +-#*() ,which get from address book - iphone

Question

In IOS,how to remove +-#*() ,which get from address book.
for example:
NSString *txt = #"[phone]"
txt = [NSString stringWithFormat:#"tel://%#", txt];
[[UIApplication sharedApplication]canOpenURL:[NSURL URLWithString:txt]];
it's a invalid number to call.
[[UIApplication sharedApplication]openURL:[NSURL URLWithString:urlToCall]]; is useless for this type number.
could I have some better options except use
[txt stringByReplacingOccurrencesOfString:#"-" withString:#""]....
I just wanna make a call.
Thanks for your help.
In fact:we can make a call by openURL: the format of call is not "tel://",It is "tel:".
So,what I should do is that:
NSString *cleanedString = [[phoneNumber componentsSeparatedByCharactersInSet:
[[NSCharacterSet characterSetWithCharactersInString:#"0123456789-+()"]
invertedSet]] componentsJoinedByString:#""];
NSURL *telURL = [NSURL URLWithString:[NSString stringWithFormat:#"tel:%#", cleanedString]];
That's OK.
————————————————————————————————
#"tel://“ isn't the right url to make a call!
we must use #"tel:"
if not some number ,as [phone] which
BOOL bCanCall = [[UIApplication sharedApplication]canOpenURL:urlToCall]; will return False. you can't make a call.

Answer 1

Try this, it may work
-(NSString *) formatIdentificationNumber:(NSString *)string
{
NSCharacterSet * invalidNumberSet = [NSCharacterSet characterSetWithCharactersInString:#"\n_!##$%^&*()[]{}'\".,<>:;|\\/?+=\t~` "];
NSString * result = #"";
NSScanner * scanner = [NSScanner scannerWithString:string];
NSString * scannerResult;
[scanner setCharactersToBeSkipped:nil];
while (![scanner isAtEnd])
{
if([scanner scanUpToCharactersFromSet:invalidNumberSet intoString:&scannerResult])
{
result = [result stringByAppendingString:scannerResult];
}
else
{
if(![scanner isAtEnd])
{
[scanner setScanLocation:[scanner scanLocation]+1];
}
}
}
return result;
}
or see the link

Answer 2

You can remove like this..
NSString *s = #"[phone]";
NSCharacterSet *unwantedStr = [NSCharacterSet characterSetWithCharactersInString:#"+()"];
s = [[s componentsSeparatedByCharactersInSet: unwantedStr] componentsJoinedByString: #""];
NSLog(#"%#", s);

Answer 3

Alternativ, some hints here, but you can optimize the code.
- (void)testExample
{
NSMutableCharacterSet *mvalidCharSet = [[NSMutableCharacterSet alloc] init];
[mvalidCharSet formUnionWithCharacterSet:[NSCharacterSet decimalDigitCharacterSet] ];
[mvalidCharSet addCharactersInString: #"+"];
NSCharacterSet *validCharSet = [mvalidCharSet copy]; // not mutable -> more efficient
NSCharacterSet *invalidCharSet = [validCharSet invertedSet];
NSString *txt = #"[phone]";
NSArray * components = [txt componentsSeparatedByCharactersInSet:invalidCharSet];
NSString * rejoin = [components componentsJoinedByString:#""];
NSLog(#"%#", rejoin);
}