Filter array by key having "version style"

Question

I'm new to php and I' still learning its logics. I have an array like this:

$arr=array("1.1"=>"Val", "1.1.1"=>"Val", "1.1.1.1"=>"Val",
       "1.1.1.2"=>"Val", "1.1.1.3"=>"Val", 
       "1.1.1.4"=>"Val", "1.1.2"=>"Val", "1.1.3"=>"Val",
       "1.1.4"=>"Val", "1.1.5"=>"Val", "1.1.6"=>"Val",
       "1.1.7"=>"Val", "1.1.8"=>"Val", "1.1.9"=>"Val",
       "1.1.10"=>"Val", "1.1.11"=>"Val");

I need to filter it extracting items of the group "1.1.1"
So, my result should be:

$arr2=array("1.1.1"=>"Val", "1.1.1.1"=>"Val",
       "1.1.1.2"=>"Val", "1.1.1.3"=>"Val", 
       "1.1.1.4"=>"Val");

I wrote the following (working) code but I was unable to put the same logic into a function usable with array_filter

$arr2=array();
$StartWith="1.1.1";
foreach ($arr as $k=>$v){
    $sp=strpos($k, $StartWith);
    if ($sp===0){
        if (strlen($k)==strlen($StartWith) ||
            strlen($k)>strlen($StartWith) && substr($k,strlen($StartWith),1)=='.'){
        $arr2[$k]=$v;
        }
    }
}

Now, my questions are:

Is it a good idea/practice to use array_filter instead of the above code?

If yes please help me writing a function (I've not understood how to).

Answer 1

array_filter() is not very useful here (even may happen that callback function make it heavior).

You can do it like below (faster and easier):-

$search = '1.1.1';

$final = [];

if(isset($arr[$search])){
    $final[$search] = $arr[$search];
 }
foreach($arr as $key=>$val){
    if(count(explode('.',$search)) < count(explode('.',$key)) && array_intersect(explode('.',$search),explode('.',$key)) == explode('.',$search)){
        $final[$key]= $val;
    }
}

print_r($final);

Output:-https://eval.in/897183