replace matches in NSString with template using NSRegularExpression - objective-c

Question

I'm trying to detect <br> or <Br> or < br>,... in NSString and replace it with \n.
I use NSRegularExpression and i wrote this code:
NSString *string = #"123 < br><br>1245; Ross <Br>Test 12<br>";
NSError *error = NULL;
NSRegularExpression *regex = [NSRegularExpression regularExpressionWithPattern:#"<[* ](br|BR|bR|Br|br)>" options:NSRegularExpressionCaseInsensitive error:&error];
NSString *modifiedString = [regex stringByReplacingMatchesInString:string options:0 range:NSMakeRange(0, [string length]) withTemplate:#"\n"];
NSLog(#"%#", modifiedString);
it works fine but it replace first matching only, not replacing all matches. Please help me to detect all matches and replace them.
Thanks

Answer 1

You currently don't handle an arbitrary amount of white space. For good measure you should also handle white space after br and also handle the closing slash since <br /> is the correct way of writing the line break in HTML.
You would end up with an pattern that looks like this
<\s*(br|BR|bR|Br|br)\s*\/*>
or written as a NSRegularExpression
NSError *error = NULL;
NSRegularExpression *regex =
[NSRegularExpression regularExpressionWithPattern:#"<\\s*(br|BR|bR|Br|br)\\s*\\/*>"
options:0
error:&error];
Edit
You could also make the pattern more compact by separating the two letters
<\s*([bB][rR])\s*\/*>

Answer 2

You're close, you need to have it handle any number of spaces after your initial <, and handle if it doesn't have any space at all.
Using your example, you can use the regex <\s*(br|BR|bR|Br|br)> to have it accept the 0 to N spaces before your BR works. You can also simplify it a little bit more by making it case insensitive with i, which allows for a cleaner looking regex to handle all the variations on BR you will see. To do that, use (?i)<\s*br>.
I think for completeness you can also include an arbitrary amount of space AFTER the br, just to handle anything that could be thrown. I agree with adding in some sort of catch for a /> to end the pattern, since <br/> is valid HTML as well. It makes the regex look a little more crazy, but it boils down to just adding the other 3 pieces.
(?i)<\s*br\s*\/?\s*>
It looks really scary, but breaks down very simply into a few parts:
(?i) turns on case insensitive to handle the variations on the br.
<\s* is the start of the tag directly followed by an arbitrary number of spaces.
br\s* is your br chars followed by an arbitrary number of spaces.
\/? is to handle 0 or 1 instances of the closing slash (to handle HTML valid tags like <br/> and <br>.
\s*> is handling an arbitrary number of spaces and then the closing >.