Detect if string contains at least 2 letters (form any language) and at least 2 words

Question

I would like to make function which detects/validates that a string got at least 2 words, and each word has at least 2 letters (except for the two letters, it can contain any other char {without numbers}, but I don't care which and how many).

Now, I am not sure if I should use regex for this or I can do it in other ways.

If I need to make regex for it, I also dont know how to do it because I need to check all the letters available.

This is the regex I got now [A-Za-z]{2,}(\s[A-Za-z]{2,}) which validates 2 words and 2 letters at least in each word.

EDIT: After re-thinking I decided to support most languages since kr-jp-cn languages work differently than rest of languages. My main rules won't let kr-jp-cn letters count as letters but as chars.

EDIT2:

This is the function I'm using based on @message answer.

function validateName($name)
{
    if (strcspn($name, '0123456789') == strlen($name)) //return the part of the string that dont contain numbers and check if equal to it length - if it equal than there are no digits - 80% faster than regex.
    {
        $parts = array_filter(explode(' ',$name)); //should be faster than regex which replace multiple spaces by single one and then explodes.
        $partsCount = count($parts);
        if ($partsCount >= 2)
        {
            $counter = 0;
            foreach ($parts as $part)
            {
                preg_match_all('/\pL/u', $part, $matches);

                if (count($matches[0]) >= 2)
                {
                    $counter++;
                }
            }
        }

        if ($counter == $partsCount)
        {
            return 'matches';
        }
    }

    return 'doesnt match';
}

Thanks for the help.

Answer 1

i would use regex also

preg_match('/\w{2,}\s+\w{2,}/u', 'word слово');

\w{2,} matching word character 2 or more. \s+ matching all spaces between and using /u unicode modifier

Edit:

I thought that such solution will help, but you need something more complex like

$text = preg_replace('/\s+/', ' ', 'word w.s');

$parts = explode(' ', $text, 2);
if (count($parts) < 2) {
    throw new \RuntimeException('Should have more than two words');
}

foreach ($parts as $part) {

    preg_match_all('/\w/u', $part, $matches);

    if (count($matches[0]) < 2) {
        throw new \RuntimeException('Should have more than two letters in word');
    }
}

Answer 2

Use Unicode character properties.

\p{L} or \p{Letter} matches a code point with the Letter property in any language. php.net documentation on Unicode character properties

Answer 3

If you are trying to use these words from the strings afterwards regex is not the way to go. Regex is not a parser. The best way I can see doing this is a combination of explode() and ctype_alpha(). Something along the lines of

$prepstring = $string;

//Remove all spaces from the original string and check that everything is a char
if(ctype_alpha(str_replace(array(' '), '', $prepstring))){

  //If everything is a char explode your string into an array
  explode($string);

  if(isset($string[1])){
    //Everything checks out, do something here.
  }

}