[SOLVED] Returning which match was found

[play_]

from this code:

$var1 = "@";
$var2 = "*";
$var3 = "/*";


$string = "Hello  world  this /* should be  ok";

$found = preg_match("/(?:\\$var1|\\$var2|\\$var3)/", $string);
if($found) {
	echo $found;
}

 

it returns true, because it find /*.

 

But is there a way to tell which one it found?

like, how would i know if it found $var1 or $var2 or $var3?

[Wildbug]

Use three different if(preg_match(... constructs (or even a case block) instead of the "|" in the pattern.

[effigy]

<pre>
<?php
$string = "Hello  world  this /* should be  ok";
$matches = array('@', '*', '/*');
foreach ($matches as &$match) {
	$match = preg_quote($match, '/');
}
$pattern = join('|', $matches);
if (preg_match("/($pattern)/", $string, $findings)) {
	echo $found, '<br>';
	print_r($findings);
}
?>
</pre>

[play_]

Thank you

[play_]

Hold on not quite.

what i'm trying to do is find out which match it found.

[Wildbug]

Can you put your variables in an array?

 

<?php
function pq($text) { return preg_quote($text,'/');}

$vars = array('*','@','/*');
$string = "Hello  world  this /* should be  ok";

preg_match('/('.implode('|',array_map("pq",$vars)).')/',$string,$match);

echo "Found \"$match[1]\"!\n";

?>

 

If not, maybe something like this:

<?php

$var1 = "@";
$var2 = "*";
$var3 = "/*";

$string = "Hello  world  this /* should be  ok";

if (is_int(strpos($string,$var1))) {
echo "Found \$var1 in \$string.";
} elseif (is_int(strpos($string,$var2))) {
echo "Found \$var2 in \$string.";
} elseif (is_int(strpos($string,$var3))) {
echo "Found \$var3 in \$string.";
} else {
echo "Nothing matched.";
}
?>

Of course, "*" will match the string since it matches the asterisk in "/*".

[obsidian]

Is this what you're after:

<?php
if (preg_match('/(@|\*|\/\*)/', $string, $match)) {
  echo "found: $match[1]";
}
?>

 

Good luck.

[effigy]

Hold on not quite.

what i'm trying to do is find out which match it found.

 

In my example it shows that "/*' was matched. What more do you need?

[play_]

effigy, your example outputs:

 

Array ( [0] => /* [1] => /* )

which comes from "print_r($findings);".

$found doesn't seem to show anything =/

 

on php 5.2.1

 

 

As for obsidian and wildbug, those worked fine. thanks.

[effigy]

Gotcha. $found must have been carried over from copying your code. There is no need for it once preg_match is placed inside an if construct. print_r was a generalization for example; I assumed you'd know how to dig into the array

[play_]

sorry :|

 

but thanks all