PHP Regex - get values between ^

Question

Cannot figure this out.

I have string

^c1_e^,^d_e^,^c^

What would be the regex to get the values between^ and ^?

I tried

preg_match("/(?<=^).*?(?=^)/", $string, $match);

which doesn't work

Answer 1

An easier solution:

<?php
$string = '^c1_e^,^d_e^,^c^';
preg_match_all('/([^^,]+)/', $string, $matches);
$strings = $matches[1];
print_r($strings); //prints an array with the elements from $string
?>

Answer 2

You need to escape the carets. These by default mean 'beginning of line' in regex:

preg_match('/(?<=\^).*?(?=\^)/', $string, $match);

The above will get the , as well, but also only the first match. If you want to avoid the commas and get all the matches, you need to match the ^ and use a capture group:

preg_match_all('/\^([^^]+)\^/', $string, $match);

The array containing the parts in between the ^ will be in the array $match[1].

A non-regex solution (though a little less robust since it cannot work with a string in a format like ^abc,def^, ^abc^ where it should give out abc,def and abc) would be to split first, then remove the ^:

$theString = '^c1_e^,^d_e^,^c^';
$elements = explode(',', $theString);
for ($i = 0; $i < sizeof($elements); $i++) {
    $elements[$i] = trim($elements[$i], "^");
}
print_r($elements);

Answer 3

You can also use this ~\^(.*?)\^~

<?php
$str="^c1_e^,^d_e^,^c^";
preg_match_all("~\^(.*?)\^~", $str,$match);
echo "<pre>";
print_r($match[1]);

OUTPUT :

Array
(
    [0] => c1_e
    [1] => d_e
    [2] => c
)