php preg_match

[paj7499]

Hi I am trying to learn regular expression and am getting some unexpected results!

 

$sentence = "This sentence contains several zzz ";

 

if(preg_match("/z{1}/",$sentence))

{

echo "Regular expression found!";

}

else

{

echo "Regular expression NOT found!";

}

 

The above code is echoing out “Regular expression found. I am confused. Shouldn't “/z{1}/” return true only if there is only one z in the string not zzz. What am I missing?

[AbraCadaver]

No it is looking for a match of z repeated 1 time and it found 1 z.  Roughly the same as [z]  (match a z character) If you want to find a z not followed by a z then something like [z][^z].  For your pattern, if you did a preg_match_all() you'd get 3 matches for 3 single z.

[.josh]

The exact pattern will vary depending on what you are actually trying to match. For example if you want to find a certain whole word, you could do

 

function isWordFound($haystack,$needle) {
  return (bool) preg_match('~\b'.$needle.'\b~',$haystack); 
}

$string = "foobar";
$word = "foo";
echo isWordFound($string,$word); // output: false

$string = "foo bar";
$word = "foo";
echo isWordFound($string,$word); // output: true

[paj7499]

Hi I am still confused

 

$subject = "This sentence contain several 'azzz' words";
$pattern = "/az{2}/";


 if (preg_match($pattern, $subject))
 {
     echo 'Regular expression <b>found!</b>';
 }
 
 else
 {
     echo 'Regular expression <b>not</b> found!';
 }
 

the above code is returning true! ('Regular expression found!)

Shouldn't preg_match  return false. My understanding is that pattern = "/az{2}/" means  match an a followed by exactly zz (2 z's not 3).  am i wrong?

[Eshe]

Your pattern matches any string with "azz" in it, followed or preceded by anything.

If you don't want it to match "azzz" you'll have to be more specific. For example:

/^azz$/ would only match the string "azz",

/azz(?!z)/ would match "azz" not followed by "z" anywhere in the string.