Question

Here is my code:

NSRegularExpression * regex;

- (void)viewDidLoad {
    NSError *error = NULL;
    regex = [NSRegularExpression regularExpressionWithPattern:@"<*>" options:NSRegularExpressionCaseInsensitive error:&error];
}

- (IBAction)findWord {  
    NSString * fileContents=[NSString stringWithContentsOfFile:[NSString stringWithFormat:@"%@/report1_index1_page1.html", [[NSBundle mainBundle] resourcePath]]];
    NSLog(@"%@",fileContents);

    NSString * modifiedString = [regex stringByReplacingMatchesInString:fileContents
                                                                options:0
                                                                  range:NSMakeRange(0, [fileContents length])
                                                           withTemplate:@"$1"];

    NSLog(@"%@",modifiedString);
}

My ‘modifiedString’ is returning (null).Why?I want to replace any characters between ‘<‘ and ‘>’ including ‘<‘ and ‘>’ simply by a space.

Answer 1

I am guessing this has a lot to do with the fact that you are assigning an autoreleased object to regex in viewDidLoad. Try adding a retain or move the line to the findWord method.

Regex

The regular expression for matching everything between < and > is incorrect. The correct way would be,

NSError *error = nil;
NSRegularExpression * regex = [NSRegularExpression regularExpressionWithPattern:@"(?<=<).*(?=>)" options:NSRegularExpressionCaseInsensitive error:&error];
if ( error ) {
    NSLog(@"%@", error);
}

Replace by space

If you want to replace the matched string with " " then you shouldn’t pass $1 as the template. Rather, use " " as the template.

NSString * modifiedString = [regex stringByReplacingMatchesInString:fileContents
                                                            options:0
                                                              range:NSMakeRange(0, [fileContents length])
                                                       withTemplate:@" "];