Count Words Function

[Anon]

I am trying to count the amount of words a user types into to a form field. This text will eventually be put into a mySQL database but I would like to get a count of the number of words the user types into the field. Is their a builtin php3 function that will do this or is this somehting that i will have to code my self.

Thanks for your help.

Ira M. Bargon III

[Anon]

Explode the string with explode(" ", $string) then use count() on the resulting array.

[[deleted]]

I wouldn't do that, because an explode around spaces would mean that
"this,that" would be counted as one word, even though there clearly are two words.

Try
echo (count(preg_split("/\w\b/",$text))-1);
that will split around \w\b, which means "a word-character followed by a word-boundry"

[kmarsh]

Aaah fancy... 🙂

[[deleted]]

"Have you had your regexp today"?

[kmarsh]

No. 😉 Still haven't learned the wonders of Regular Expressions. Please teach me...

Simple one... allow 3-20 alphanumeric characters and spaces...

[[deleted]]

if (preg_match("/^[0-9a-zA-Z ]{3,20}$/",$string))
{
echo "yes";
}
else
{
echo "no";
};

/^{[0-9a-zA-Z]{3,20}$/}
means:
^ - must be the beginning of the string
[ ] - whatever is between these brackets must match
0-9 - any number from 0 to 9
a-z - any letter from a to z (in lower case)
A-Z - any letter from A to Z (in upper case)
{3,20} - whatever was directly in front of this must match at least 3 and at most 20 times.
$ - must be the end of the string

[kmarsh]

Thanks vincent! For never being in the dark about regular expressions, I am much obliged 😉

[spider_man]

That's just too gay.

for ($i=0; $i<strlen($string); $i++)

if (
(
(ord($string[$i-1])<65) ||
(ord($string[$i-1])>90 && ord($string[$i-1])<97) ||
(ord($string[$i-1])>122)
)
&&
(

ord($string[$i])>64 && ord($string[$i])<91) ||
ord($string[$i])>96 && ord($string[$i])<123)
)
)
$n_words++;

echo 'You've got ' . $n_words ' words, baby!';

[ccgorman]

Right,

It looks like you know what you are talking about. Here is my situation. If you have a page with HTML elements, and you want to display 500 words at a time, but you don't want to cut off a tag (ie, if there is a <a href="..."> and you want to make sure that you include the </a>. How would you check if there is a missing closing element and increment the number of words to be displayed up or down so that this element is printed to the page.

Any thoughts?

-David

[spider_man]

Well, the algorithm increments the counter when it finds a char that's not an uppercase or lowercase letter followed by another that IS a letter.

I don't get one thing, you want to output html code, or plaint text?
Assuming the second one, you could first filter the text with the striptags function provided by php.
Then you could put the snippet I've sent inside a for loop and end it when $n_words reaches 500, while you print it.

Hope I've been clear, please let me know.

[spider_man]

Oops, thought you were talkin' to me.

[[deleted]]

Wohee, that's a cool way of doing things the hard way, if you don't count the 4 syntax errors....

[[deleted]]

Grabbing the words is easy, but looking for closing tags is a different matter.
For one, how would you decide which tags 'must' be closed, and which tags can be left open? You mention the HREF tag, which obviously should be closed.
But what about the B tag, or TD? font?

If you'd want to go through with it, a simple way would be to check your text for HTML tags. every time you find a new opening tag, add it to an array of tags.
Whenever you find a closing tag, check if the last tag in the 'opened' array is of the same type. If so, that tag must be closed now, and you remove that opening tag from the array (pop the little bastard off the end of the array, making the array shorter)
When the array is empty, your document should be complete.

There's also a regexp example of finding matching tags somewhere in there... but I forgot where. :-)

[Anon]

Not to mention multiple spaces together