Search for array keys with a pattern

Question

Let's say I have the following array:

$arr = array( "number2"=>"valid", "number13"=>"valid" ); 

and I need to find if the key exists with number* .

For $arr this will be true. For the following array:

 $arr2 = array( "key"=>"foo", "key2"=>"foo2" ); 

this will return false.

Answer 1

This means that the number should be followed by the actual number (edit: or nothing at all), if necessary, adjust the regular expression. For example, everything starting with "number", you can use /^number/ .

 if(count(preg_grep('/^number[\d]*/', array_keys($arr))) > 0) { return true; } else { return false; } 

Answer 2

Use regex.

 foreach ($arr as $key => $value) { // NOTE: check for the right format of the regular expression if (preg_match("/^number([0-9]*)$", $key)) { echo "A match was found."; } else { echo "A match was not found."; } } 

Answer 3

Here is a simple function that will do what you want:

 function preg_grep_key($pattern, $input) { return preg_grep($pattern, array_keys($input)); } // ----- Usage ----- $arr = array( "number2"=>"valid", "number13"=>"valid" ); if (count(preg_grep_key('/^number/', $arr)) === 0) { // Nope } else { // Yep }