Position of a character in a NSString or NSMutableString

Question

I have searched for hours now and haven\'t found a solution for my problem. I have a NSString which looks like the following:

\"spacer\": [\"value1\", \"valu

Answer 1

To find occurrences of a string within a string, use the rangeOfXXX methods in the NSString class. Then you can construct NSRanges to extract substrings, etc.

This example removes only the first set of open/close brackets in your sample string...

NSString *original = @"\"spacer\": \[\"value1\", \"value2\"], \"spacer\": \[\"value1\", \"value2\"]";
NSLog(@"%@", original);

NSRange startRange = [original rangeOfString:@"\["];
NSRange endRange = [original rangeOfString:@"]"];

NSRange searchRange = NSMakeRange(0, endRange.location);
NSString *noBrackets = [original stringByReplacingOccurrencesOfString:@"\[" withString:@"" options:0 range:searchRange];
noBrackets = [noBrackets stringByReplacingOccurrencesOfString:@"]" withString:@"" options:0 range:searchRange];
NSLog(@"{%@}", noBrackets);

The String Programming Guide has more details.
You might alternatively also be able to use the NSScanner class.

Answer 2

This worked for me to extract everything to the index of a character, in this case '[':

NSString *original = @"\"spacer\": \[\"value1\", \"value2\"], \"spacer\": \[\"value1\", \"value2\"]";
NSRange range = [original rangeOfString:@"\["];
NSString *toBracket = [NSString stringWithString :[original substringToIndex:range.location] ];
NSLog(@"toBracket: %@", toBracket);