How do I assign different variables for an SQL result that is joined

I am working on a project and I came across something that I just don’t know how to solve. I have 3 tables in a Firebird database :

|TEAM | ACTIVITY|
|-----|---------|
| 1   | JUMPING |

Team 1 – jumping

    | PLAYER | TEAM |
    |--------|------|
    |   10   |   1  |
    |   15   |   1  |

The player 10 and 15 are in Team 1

And the last table is this one :

    | PLAYER_ID | PLAYER_NAME | AGE | HEIGHT |
    |-----------|-------------|-----|--------|
    |     10    |     Alex    |  25 |   175  |
    |     15    |    Andrew   |  17 |   187  |

Player 10 -> Alex -> age 25 -> height 175

Player 15 -> Andrew -> age 17 -> height 187

My main goal is to make an HTML page with a textbox, in the textbox I will type the Team number (in this case I will type 1) and it will show me who is in the team with his qualities (age and height).

    $teamPlayers= "SELECT PLAYER_ID
             FROM TEAM
             WHERE TEAM= $teamNumber;";

( keep in mind, $teamNumber is the value introduced by the user in the HTML textbox )

    $y = ibase_query($dbConnect, $teamPlayers);
    while ($row1 = ibase_fetch_object($y)) {
    $array1 = get_object_vars($row1);
    echo $array1["PLAYER_ID"];} 

This program outputs 1015 which is fine, means the program is working, but I need to know if there is other way of selecting the players of the team, since this one basically joins their IDs together, and I need them separately so I can assign their IDs to separate variables.

Answer

I believe you need to iterate the array you get from SQL database. Ex.

$teamPlayers= "SELECT PLAYER_ID
         FROM TEAM
         WHERE TEAM= $teamNumber;"
$playerIDArray = array();
$ct = 0;
foreach ($teamPlayers as $players) {
     $ct += 1;
     $playerIDArray[$ct] = $players["PLAYER_ID"]
}