php function return variable not working

Question

I have created a function(recursive). It will call again if it has parent id otherwise it will return last Id. The Problem is I'm unable to return the last Id which passed in the Function argument. It is showing complete if I use var_dump() or echo but unable to return in return().

Here is Sample Code :

function abc($dbcon,$id){
    $get_quy = "select * from member_mst where member_status=0 and pid=".$id;
    $get_rs = $dbcon->query($get_quy);

    $get_numrows = mysqli_num_rows($get_rs);
    $get_rel = mysqli_fetch_assoc($get_rs);

    if($get_numrows>0){
        abc($dbcon,$get_rel['member_id']);
    }else{
        //var_dump($id);//This will return => "String '3'" 
        return $id;
    }
}
var_dump(abc($dbcon,1));//Returning Null

i think it should be `return abc($dbcon,$get_rel['member_id']);`. You execute the function but never get results from the recursion calls — Kaddath, May 28 '18 at 07:10
See http://php.net/manual/mysqli.quickstart.prepared-statements.php — Phil, May 28 '18 at 07:11
@abhichavda What results do you get? Can you add: `print $get_numrows."\n"` before `if($get_numrows>0){`? — user4035, May 28 '18 at 07:19
@Kaddath It is worked. I have used return while calling the function again. `return abc($dbcon,$get_rel['member_id']);` — abhi chavda, May 28 '18 at 07:20

score 0 · Accepted Answer · answered May 28 '18 at 07:28

Here is the working code. The Problem was I wasn't using return() while calling the function again.

function abc($dbcon,$id){
    $get_quy = "select * from member_mst where member_status=0 and pid=".$id;
    $get_rs = $dbcon->query($get_quy);

    $get_numrows = mysqli_num_rows($get_rs);
    $get_rel = mysqli_fetch_assoc($get_rs);

    if($get_numrows>0){
        return abc($dbcon,$get_rel['member_id']); // Updated Code Here
    }else{
        return $id;
    }
}
var_dump(abc($dbcon,1));//Returning "3"

php function return variable not working

1 Answers1