Regular expressions. preg_replace

Question

Good day. The beginner in regular expressions, but I want to understand. There is a string, it needs to be reduced to one type, i.e. leave only the name. Options: http://vk.com/id62334653 vk.com/id62334653 id62334653, etc.

When working with value:

$data = 'http://vk.com/id62334653'; $pattern = '/((http(s)?:\/\/)?(new\.)?(m\.)?(vk\.com\/)?)([a-z_0-9.]*)?(.*)?/ius'; $data = preg_replace($pattern, '[$7]', trim($data)); echo '<br />'.$data2.'='.$data;

No problem, the result: http://vk.com/id62334653=[id62334653]

But if the string:

 $data = 'Михаил Успенский http://vk.com/m.uspenskiy';

result: Mikhail Uspensky [m.uspenskiy]

They advised to do this:

 $pattern = '/((http(s)?:\/\/)?(new\.)?(m\.)?(vk\.com\/)?)([a-z_0-9.]*)?(.*)?/ius'; $data = preg_replace($pattern, '[$7]', trim($data)); echo '<br />'.$data2.'='.$data;

But the problem is that a blank line is issued here if we remove the "?" after (http (s)?: //), then everything is fine, but then expressions like "vk.com/m.uspenskiy" do not pass, because there is no "http" in it, but this input option is also possible.

A question how to consider all this data? those. How to consider the options: "vk.com/m.uspenskiy" and 'Mikhail Uspensky http://vk.com/m.uspenskiy ', so that the result is "m.uspenskiy"?

because your pattern coincides with vk.com/m.uspenskiy 'and $ 7 corresponds to the standard m.uspenskiy, so the url is replaced.
And "Mikhail Uspensky" is not added anywhere, but simply is present in the original line
To make it easier to deal with regulars, I advise regex101.com , especially there is the item "regex debugger" - shows step by step how the handler understood your pattern.

Answer 1 · 2016-08-16T08:31:04

You find in the full line ( Михаил Успенский http://vk.com/m.uspenskiy ) a substring on the specified pattern ( http://vk.com/m.uspenskiy ) and replace it with its part ( [m.uspenskiy] ). So it turns out.

If you need to get the desired part in a single variable, you need to use something like this:

 $data = 'Михаил Успенский http://vk.com/m.uspenskiy'; $pattern = '/((http(s)?:\/\/)(new\.)?(m\.)?(vk\.com\/)?)(?P<id>.*)?/ius'; preg_match($pattern, $data, $matches); $id = isset($matches['id']) ? $matches['id'] : ''; echo '<br />'.$data.'='.$id;

preg_match roughly speaking, parses the input string into parts according to the pattern and you can get each found part separately.

Notice that I added ?P<id> inside the brackets in the part of the regular season I want to see in $matches['id'] . This allows you to safely change the regular schedule without being tied to the group index.

Thank you, but "Mikhail Uspensky vk.com/m.uspenskiy " for "$ pattern = '/ ((http (s)?: \ / \ /)? (New \.)? (M \.)? (Vk \ .com \ /)?) ([a-z_0-9.] *)? (. *)? / ius'; "
not how to consider the options: "vk.com/m.uspenskiy" and 'Mikhail Uspensky vk.com/m.uspenskiy '
If you change the regular season - of course, it does not fit.
or slightly modify the regular schedule to the universal version.

Regular expressions. preg_replace

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