Regex to replace asterisk characters with html bold tag

Question

Does anyone have a good regex to do this? For example:

This is *an* example

should become

This is <b>an</b> example

I need to run this in Objective C, but I can probably work that bit out on my own. It's the regex that's giving me trouble (so rusty...). Here's what I have so far:

s/\*([0-9a-zA-Z ])\*/<b>$1<\/b>/g

But it doesn't seem to be working. Any ideas? Thanks :)

EDIT: Thanks for the answer :) If anyone is wondering what this looks like in Objective-C, using RegexKitLite:

NSString *textWithBoldTags = [inputText stringByReplacingOccurrencesOfRegex:@"\\*([0-9a-zA-Z ]+?)\\*" withString:@"<b>$1<\\/b>"];

EDIT AGAIN: Actually, to encompass more characters for bolding I changed it to this:

NSString *textWithBoldTags = [inputText stringByReplacingOccurrencesOfRegex:@"\\*([^\\*]+?)\\*" withString:@"<b>$1<\\/b>"];

Answer 1

Why don't you just do \*([^*]+)\* and replace it with <b>$1</b> ?

Answer 2

You're only matching one character between the *s. Try this:

s/\*([0-9a-zA-Z ]*?)\*/<b>$1<\/b>/g

or to ensure there's at least one character between the *s:

s/\*([0-9a-zA-Z ]+?)\*/<b>$1<\/b>/g

Answer 3

I wrote a slightly more complex version that ensures the asterisk is always at the boundary so it ignores hanging star characters:

/\*([^\s][^\*]+?[^\s])\*/

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Test phrases with which it works and doesn't:

Answer 4

This one regexp works for me (JavaScript)

x.match(/\B\*[^*]+\*\B/g)