Php form update in update the form

Question

I am running while loop and fetch 3 records from database. and then update it on same page. Every record have submit button. But after edit when i submit the form it catchs the values of last record only and update other rows with the last record values. Please if somebody help me out i'll be very thankful. Remember it catches the exact (id) but the other parameters are only of last row.

<form method="post" action="">
    <table width="700" border="1">
       <tr><th><?php echo $_SESSION['teamtwo']; ?></th></tr>
       <tr>
         <th>Player Name</th>
         <th>Runs</th>
         <th>Edit</th>
         <th>Save</th>
       </tr>
       <?php
        $team = new DBConnection();
        $condition = "WHERE teamname = '".$_SESSION['teamtwo']."' and datecreated = CURDATE()";
        $sel_player = $team->SelectRecord(array("*"),"`match`","$condition");
        //$sel_player = mysql_query("SELECT * FROM `match` WHERE teamname = '$team1' and datecreated = CURDATE()") or die(mysql_error());
        while($get_player = mysql_fetch_array($sel_player))
        {
        $totalruns = $get_player['runs_bat'];
        $totalballs = $get_player['ball_bat'];
        @$strike = $totalruns / $totalballs * 100; 
        ?>
        <tr>
          <td><input type="text" name="player_name" value="<?php echo $get_player['player_name']; ?>" disabled="disabled" /></td>
           <td><input type="text" name="runs" value="<?php echo $get_player['runs_bat']; ?>" size="1" /></td>

           <td><button><a href="?player=<?php echo $get_player['id']; ?>">Edit</a></button></td>
           <td><input type="submit" value="Save" name="team" /></td>
             </tr>
         <?php 
        } ?>
     </table>
   </form>
<?php } ?>
 </div>
 </div>
</body>
</html>
<?php

    if(isset($_POST['team'])){
        $runs = $_POST['runs'];
        $balls = $_POST['ball'];




        $object = new DBConnection();
        $arr_Field=array("runs_bat","ball_bat","player_status","how_out","opposite_bowl","opposite_player","sr","overs","bowl_ball","runs_ball","extra","madien");
        $arr_Values=array("$runs","$balls","$status","$how_out","$opposite_bowler","$opposite_player","$sr","$over","$bowls","$score","$extra","$madien");
        $condition = "WHERE id = '".$_REQUEST['player']."'";
        //echo $_REQUEST['player'];

        //echo $runs.$balls;

        $object->UpdateRecord("`match`",$arr_Field,$arr_Values,"$condition") or die(mysql_error());
        //header("Location:extra.php?update");


    }

Answer 1

the problem is you are having one form and when you submit the form it will submit the last rows values because you are having same name for all 3 rows inside 1 form.

Solution:-

Create form element inside the while loop and close it inside the while loop itself . Like this you will have 3 forms each for 3 rows.

Code Example:-

    while($get_player = mysql_fetch_array($sel_player))
    {
    $totalruns = $get_player['runs_bat'];
    $totalballs = $get_player['ball_bat'];
    @$strike = $totalruns / $totalballs * 100; 
    ?>
    <form>
    <tr>
      <td><input type="text" name="player_name" value="<?php echo $get_player['player_name']; ?>" disabled="disabled" /></td>
       <td><input type="text" name="runs" value="<?php echo $get_player['runs_bat']; ?>" size="1" /></td>

       <td><button><a href="?player=<?php echo $get_player['id']; ?>">Edit</a></button></td>
       <td><input type="submit" value="Save" name="team" /></td>
    </tr>
    </form>
     <?php 
    } ?>