Count all word including numbers in a php string

Question

To count words in a php string usually we can use str_word_count but I think not always a good solution

good example:

$var =\"Hello world!\";
echo str_         


        

Answer 1

The most wide-spread method of counting words in a string is by splitting with any kind of whitespace:

count(preg_split('~\s+~u', trim($text)))

Here, '~\s+~u' splits the whole text with any 1 or more Unicode whitespace characters.

The disadvantage is that !! is considered a word.

In case you want to count letter and number words (i.e. strings of text that are only made up of just letters or just numbers) you should consider a preg_match_all like

if (preg_match_all('~[-+]?[0-9]*\.?[0-9]+(?:[eE][-+]?[0-9]+)?|\d+|(?>\p{L}\p{M}*+)+~u', $text, $matches)) {
    return count($matches[0]);
}

See the regex demo and the PHP demo:

$re = '~[-+]?[0-9]*\.?[0-9]+(?:[eE][-+]?[0-9]+)?|\d+|(?>\p{L}\p{M}*+)+~u';
$text = "The example number 2 is a bad example it will not \ncount numbers  and punctuations !! X is 2.5674.";
if (preg_match_all($re, $text, $matches)) {
    echo count($matches[0]);
} // 18 in this string

The [-+]?[0-9]*\.?[0-9]+(?:[eE][-+]?[0-9]+)? regex is a well-known integer or float number regex, and (?>\p{L}\p{M}*+)+ matches any 1 or more letters (\p{L}), each of which can be followed with any amount of diacritic marks (\p{M}*+).

Regex details

• [-+]?[0-9]*\.?[0-9]+(?:[eE][-+]?[0-9]+)? - an optional - or +, 0+ ASCII digits, an optional ., 1+ ASCII digits, an optional sequence of e or E, an optional - or + and then 1+ ASCII digits
• | - or
• \d+ - any 1 or more Unicode digits
• | - or
• (?>\p{L}\p{M}*+)+ - 1 or more occurrences of any Unicode letter followed with any 0+ diacritic symbols.

In case you just want to count text chunks consisting of solely digits and letters (with diacritics) mixed up in any order, you may also use

'~[\p{N}\p{L}\p{M}]+~u'

See another regex demo, \p{M} matches diacritics, \p{N} matches digits and \p{L} matches letters.