计算php字符串中包括数字在内的所有单词

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:

$var ="Hello world!";
echo str_word_count($str);
print_r(str_word_count($str, 1));

-->output

   2
   Array ( [0] => Hello [1] => world ) 

bad example:

$var ="The example number 2 is a bad example it will not 
count numbers  and punctuations !!";

-->output:

  14
  Array ( [0] => The [1] => example [2] => number [3] => is [4] => a
  [5] => bad [6] => example [7] => it [8] => will [9] => not 
  [10] => count [11] => numbers [12] => and [13] => punctuations ) 

Is there a good predefined function to do this properly or do I have to use preg_match() ?

Answer 1

You can always split your string by whitespace and count the results:

$res = preg_split('/\s+/', $input);
$count = count($res);

With your string

"The example number 2 is a bad example it will not 
count numbers  and punctuations !!"

This code will produce 16 .

The advantage of using this over explode(' ', $string) is that it will work on multi-line strings as well as tabs, not just spaces. The disadvantage is that it's slower.

Answer 2

The following using count() and explode() , will echo:

The number 1 in this line will counted and it contains the following count 8

PHP:

<?php

$text = "The number 1 in this line will counted";

$count = count(explode(" ", $text));

echo "$text and it contains the following count $count";

?>

---

Edit:

Sidenote:
The regex can be modified to accept other characters that are not included in the standard set.

<?php

$text = "The numbers   1  3 spaces and punctuations will not be counted !! . . ";

$text = trim(preg_replace('/[^A-Za-z0-9\-]/', ' ', $text));

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


// used for the function to echo the line of text
$string = $text;

    function clean($string) {

       return preg_replace('/[^A-Za-z0-9\-]/', ' ', $string);

    }

echo clean($string);

echo "<br>";

echo "There are ";
echo $count = count(explode(" ", $text));
echo " words in this line, this includes the number(s).";

echo "<br>";

echo "It will not count punctuations.";

?>

Answer 3

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 (ie 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.

Answer 4

使用count(explode(' ', $var));

Answer 5

You can try this,

<?php
function word_count($sentence)
{
$break = explode(" ",$sentence);
$count = count($break);
return $count;
}
$count =  "Count all words of this sentence";
echo word_count($count); 
//Output 6
?>

Here is more about Word Count In PHP

Answer 6

You may also use the below code it's working for me.

    function get_num_of_words($string) {
        $string = preg_replace('/\s+/', ' ', trim($string));
        $words = explode(" ", $string);
        return count($words);
    }

    $string="php string word count in simple way";
    echo $count=get_num_of_words($string);

The result will be 7

Answer 7

I know the question is old, Still im sharing the fix i have adopt for this.

$str ="Hello world !";
// you can include allowed special characters  as third param.
print_r(str_word_count($str, 1, '!'));

code output is

Array ( [0] => Hello [1] => world [2] => ! )

if you want to include more words u can specify as third param.

print_r(str_word_count($str, 1, '0..9.~!@#$%^&*()-_=+{}[]\|;:?/<>.,'));

from 0..9. will include all numbes, and other special characters are inserted individually.

Answer 8

Just some improvement your solution

function stringWordNumberCount($text){
    if (!$text) {
        return 0;
    }

    //Clean the text to remove special character
    $text = trim(preg_replace('/[^A-Za-z0-9\-]/', ' ', $text));

    //Remove continus space on text
    $text = trim( preg_replace('/\s+/', ' ',$text));

    //count space
    return count(explode(' ', $text));

}

Answer 9

ans:

function limit_text($text, $limit) {
    if(str_word_count($text, 0) > $limit) {
        $words = str_word_count($text, 2);
        $pos = array_keys($words);
        $text = substr($text, 0, $pos[$limit]) . '...';
    }
    return $text;
}