Agfdhyk

I am trying to list players into a page, the page already uses a sql query, so i have put this query in a function (below) which sitting in the functions.php, in another file teampage.php I have called the function <?php echo listGoalkeepersForTeam(); ?> , but it is throwing a Notice: Undefined variable: id in C:xampp***functions.php on line 77

how to get function to read the teampage.php?id=1 and display the information accordingly.

function listGoalkeepersForTeam() {



    $dbServername = "****";

    $dbUserrname = "****";

    $dbPassword = "****";

    $dbName = "****";



    $sql  = "select player.fname as fname, player.sname as sname, player.position as position, player.profile_img as profile_img from player_team 

    join player on player_team.player_id = player.player_id 

    join team on player_team.team_id = team.team_id 

    where team.team_id = '$id'

    AND player.position = 'Goalkeeper'";





    $conn = mysqli_connect($dbServername, $dbUserrname, $dbPassword, $dbName);

    $query = mysqli_query($conn, $sql);



    if (!$query) {

        die ('SQL Error: ' . mysqli_error($conn));

    }

    ?>

       <?php

       $no  = 1;

       $total   = 0;

       while ($row = mysqli_fetch_array($query))

       { ?>



                <img class="rounded-circle img-fluid d-block mx-auto mr-4" src="img/db_img/players/child.png" alt="<?php echo ($row['fname']. " " .$row['sname']) ?>">

                <h3><?php echo ($row['fname']. " " .$row['sname']) ?><br>

                <small><?php echo ($row['position']) ?></small>

                </h3>





<?php      $no++;

       } 



}

You are not getting the id from GET variable.

function listGoalkeepersForTeam() {

//insert line

$id = filter_input(INPUT_GET, 'id');

....