[RESOLVED] Reg ex vowel check

[sssphp]

I want to check a text string for words that do not contain a vowel. I've tried the following, but something's not right:

if ( preg_match('/\b[^{aeiouy]+\b/i',} $text) )
{ // there is a word with no vowel }
else
{ // all words have at least 1 vowel }

if ( preg_match('/\b[^{a|e|i|o|u|y]+\b/i',} $text) )
{ // there is a word with no vowel }
else
{ // all words have at least 1 vowel }

[rkolbe]

if ( preg_match('/[^{a|e|i|o|u|y]+/i',} $text) )
{ // there is a word with no vowel }
else
{ // all words have at least 1 vowel }

I dont think you need \b...I usually just use / for boundries. That should work though.

[laserlight]

From what I see, the problem is that the word boundaries are also matching the space in between words, and this space certainly is not a vowel. One possible solution could be to include whitespace in the character class:

if (preg_match('/\b[^aeiouy\s]+\b/i', $text))
{
    // there is a word with a vowel
}
else
{
    // all words have at least one vowel
}

I dont think you need \b...I usually just use / for boundries.

The \b is a word boundary, not a pattern delimiter.

[bpat1434]

Here's one that works. It's reverse of what you initially have, but it works.

$pattern = '/\b[a|e|i|o|u|y]+\b/i';

Now, with words like "cwm" or "Th1s h4s n0 v0w3ls 😉" it goes to the else statement. Otherwise, it has vowels.

if(preg_match($pattern, $text))
  // Has vowels
else
  // Has no vowels

Or you can reverse it if needed:

if(!preg_match($pattern, $text))
  // Has no vowels
else
  // Has vowels

Hope that helps.

[laserlight]

Did you test that, bpat1434? It doesnt seem to work for me.

[bpat1434]

Yeah, tested it, works for me. PHP 5.....

and just to prove it:
Check it out here

[laserlight]

Very interesting. It seems to work with your test data, but not with mine :o

Test with:

$strings = array(
    'git bzr',
    ' git bzr',
    'git bzr ',
    "git\tbzr",
    'hg bzr',
    ' hg bzr',
    'hg bzr ',
    "hg\tbzr",
    'git monotone',
    'git monotone ',
    ' git monotone',
    "git\tmonotone");

EDIT:
Okay, mine works. The thing is, it should not be:
There were no vowels found in any word(s) in the string.
It should be:
There were no vowels found in some word in the string.

Likewise,
There were vowels found somewhere in the string.
should be:
There were vowels found in all words in the string.

[NogDog]

If you want to know each of the words that does not have a vowel, you'd want to use preg_match_all:

$test = "The quick brown fox jumped over the lazy dog. Th qck brwn fx jmpd vr th lz dg.";
if(preg_match_all('/\b[^aeiouy\s]+\b/i', $test, $matches))
{
   echo "<p>Found the following words with no vowels:</p>\n<ul>\n";
   foreach($matches[0] as $word)
   {
      echo "<li>$word</li>\n";
   }
   echo "</ul>\n";
}
else
{
   echo "<p>All the words had vowels.</p>\n";
}

[sssphp]

laserlight wrote:

From what I see, the problem is that the word boundaries are also matching the space in between words, and this space certainly is not a vowel.

Thanks, Laserlight -- you hit the nail on the head. I feel better now -- I thought I was missing something dumb, but that was a pretty astute observation. :p

You're also right about Bpat's idea, but thanks for that idea too.

[Weedpacket]

No....... \b does not match the space character, it's a zero-width assertion (matching a space character would be \w). What is matching the spaces is the [^aeiouy].

/\b[^aeiouy]+?\b/i

fixes that, but note that it will also match gaps between words (since they don't contain vowels).

It might be simpler to preg_split('/\w+/', $the_string) and then look at the words that result (because, among other things, \b matches either side of the "'" in "don't", so that would count as two words - one of them without vowels).

[laserlight]

No....... \b does not match the space character, it's a zero-width assertion (matching a space character would be \w). What is matching the spaces is the [^aeiouy].

Oops, rather loose talk there, I see. Yeah, that's what I meant, i.e., the original pattern is matching the whitespace between the word boundaries.

[sssphp]

Thanks for that catch, Weedpacket.

I think the simplest solution is to drop the word boundaries altogether, and use \s to define words; then add an apostrophe to the list of no-match chars, like so:

if ( preg_match('/\s[^{aeiouy\']+\s/',} $string))
{ echo 'there is a word with no vowel'; }
else { echo 'all words have at least 1 vowel'; }

That seems to work in testing...until someone says different :o

[laserlight]

Well, what about a word containing a dash? They might be confused with two words separated by a dash, and it is very difficult to differentiate between them.

[sssphp]

Actually, I think I'll use the following, so numbers and emoticon 'words' won't get flagged:

if ( preg_match('/\s[^{aeiouy\W\d]+\s/',} $string))
{ echo 'there is a word with no vowel'; }
else { echo 'all words have at least 1 vowel'; }

[sssphp]

Don't all hyphenated words have vowels on both sides of the hyphen? If so, then it's not a problem...have to think about that. Anyway, the last solution should cover it.

[Weedpacket]

Don't all hyphenated words have vowels on both sides of the hyphen?

What, like "Coca-Cola" or "self-confident" or "one-quarter Irish" (as opposed, of course to "one quarter-Irish")?