Index of a character from the NSString in iOS - ios

Question

I have an NSString for example "This is my question".I want to find all the indices of the character/substring "i" ie In this case If index starts from 0,then I want 2,5,16 as my answer.

Answer 1

The other answer is a bit of an overkill. Why don't you simply iterate over the characters like this:
NSString *x = #"This is my question";
for (NSUInteger i=0;i<[x length];i++)
{
if ([x characterAtIndex:i]=='i')
{
NSLog(#"found: %d", i);
}
}
It outputs exactly your positions:
found: 2
found: 5
found: 16

Answer 2

I'd like suggest my solution. It is like this:
NSString* str = #"This is my question";
NSArray* arr = [str componentsSeparatedByString: #"i"];
NSMutableArray* marr = [NSMutableArray arr];
NSInteger cnt = 0;
for (NSInteger i = 0; i < ([arr count]); i++)
{
NSString* s = [arr objectAtIndex: i];
cnt += [s length];
[marr addObject: [NSNumber numberWithInt: cnt]];
cnt += [#"i" length];
}
NSLog(#"%#", [marr description]);
On console:
2
5
16

Answer 3

I don't know is there any built-in functions available for doing this. You can use this method:
- (NSMutableArray *)indexOfCharacter:(char)c inString:(NSString*)string
{
NSMutableArray *returnArray = [[NSMutableArray alloc] init];
for(int i=0;i<string.length;i++)
{
if(c == [string characterAtIndex:i])
{
[returnArray addObject:[NSNumber numberWithInt:i]];
}
}
return returnArray;
}

Answer 4

Using NSRange and loop and with some string manipulation you can easily do it.
NSString *string = #"This is my question";
NSString *substring = #"i";
NSRange searchRange = NSMakeRange(0,string.length);
NSRange foundRange;
while (searchRange.location < string.length)
{
searchRange.length = string.length-searchRange.location;
foundRange = [string rangeOfString:substring options:nil range:searchRange];
if (foundRange.location != NSNotFound)
{
// found an occurrence of the char
searchRange.location = foundRange.location+foundRange.length;
NSLog(#"Location of '%#' is %d",substring,searchRange.location-1);
}
}
EDIT
Using NSRegularExpression and NSRange you can do like this.
NSString *string = #"This is my question";
NSString *substring = #"i";
NSRegularExpression *regex = [NSRegularExpression regularExpressionWithPattern:substring
options:0
error:NULL];
[regex enumerateMatchesInString:string options:0 range:NSMakeRange(0, [string length])
usingBlock:^(NSTextCheckingResult *result, NSMatchingFlags flags, BOOL *stop) {
NSRange range = [result range];
NSLog(#"Location of '%#' is %d",substring, range.location);
}];
output is
Location of 'i' is 2
Location of 'i' is 5
Location of 'i' is 16

Answer 5

This is my attempt at a no loop code of getting what you want. I coded this blind, meaning not-tested etc. Its basically a recursive function, but I think it gets you the general idea.
- (NSArray *)getAllEyes:(NSString *)s index:(int)index) {
if (!s || s.length <= 0 || index >= s.length) return [NSArray new];
NSRange *r = [s rangeOfString(#"i") options:NSLiteralSearch range:NSMakeRange(index, s.length - index)];
if (r.location == NSNotFound) {
return [NSArray new];
} else {
NSMutableArray *array = [NSMutableArray new];
[array addObject:#(r.location)];
[array addObjectsFromArray:[self getAllEyes:s index:r.location + 1]];
return array;
}
}
// usage:
NSArray *allEyes = [self getAllEyes:#""];
for (NSNumber *n in allEyes) {
NSLog(#"i = %#", n);
}