![]() You can repeat all the allowed characters before and after matching /products/ using the same optional character class. PHP regex extract url with pattern from string That will extract the first URL in any string (of course, this does no error checking, so if the string really doesn't contain any URL's it won't work, but take a look at the NSRegularExpression class to see how to get around it. NSString *match = [someString substringWithRange:[expression rangeOfFirstMatchInString:someString options:NSMatchingCompleted range:NSMakeRange(0, [someString match) // Correctly prints '' NSString *someString = is a sample of a sentence with a URL within it." The code you want essentially looks like this (using John Gruber's super URL regex): NSRegularExpression *expression = For a tutorial on using the class, see here. ![]() You can construct a simple one to extract the URL using the NSRegularExpression class, or find one online that you can use. Far less prone to error than regular expressions. Var text = 'Find me at and also at ' var html = urlify(text) Ĭonsole.log(html) How can I extract a URL from a sentence that is in a NSString?Įdit: I'm going to go out on a limb here and say you should probably use NSDataDetector as Dave mentions. Using preg_replace() as mentioned in the other answers, using the following regex should be one of the most accurate, if not the most accurate, method for detecting a link: (?i)\b((?: :(?:/ John Gruber has spent a fair amount of time perfecting the "one regex to rule them all" for link detection. > print re.findall(r'(https?:// )', myString) If there can be multiple links you can use something similar to below > myString = "These are the links and " ![]() There may be few ways to do this but the cleanest would be to use regex > myString = "This is a link " How do you extract a url from a string using python? You need to repeat your find function to match the next one and use the new group array. You should try to see if there is something in m.group(0), or surround all your pattern with parenthesis and use m.group(1) again. M.group(1) gives you the first matching group, that is to say the first capturing parenthesis. ![]()
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