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");


    }

score 2 · Accepted Answer · answered Oct 13 '12 at 10:36

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 
    } ?>

score 0 · Answer 2 · answered Oct 13 '12 at 10:36

1. you need to make input array in while because name attribute is overwriting in loop

<td><input type="text" name="player_name[<?php echo $get_player['id']?>]" value="<?php echo $get_player['player_name']; ?>" disabled="disabled" /></td>
<td><input type="text" name="runs[<?php echo $get_player['id']?>]" value="<?php echo $get_player['runs_bat']; ?>" size="1" /></td>

2. you have all text boxes mean if press submit button of one row, then also you will get all textboxes as php side so make hidden variable in form to get which button clicked

//write javascript in your page
<script>
function setPlayerId(id) {
  document.getElementById('playerid').value=id;
}
</script>

//take hidden field into form
<input type='hidden' name='playerid' value='0'>

//write down onlick button event

<input type="submit" value="Save" name="team" onClick="setPlayerId('<?php <?php echo $get_player['id']?>?>')"/>

3. Now in php you will get that as below

echo $_POST['player_name'][$_POST['playerid']];

// same way you can do your insert or update.

score 0 · Answer 3 · answered Oct 13 '12 at 10:38

this code must work

<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 $get_player['player_name']; ?>" disabled="disabled" /></td>
           <td><input type="text" name="runs<?=$get_player['id']?>" value="<?php echo $get_player['runs_bat']; ?>" size="1" /></td>
// you didnt write this i added
<input type="text" name="ball<?=$get_player['id']?>" value="<?php echo $get_player['ball_bat']; ?>" size="1" />

           <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'])){
    $runsname = 'runs'.$_GET['player'];
    $ballsname = 'ball'.$_GET['player'];
        $runs = $_POST[$runsname];
        $balls = $_POST[$ballsname];




        $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");


    }

Php form update in update the form

3 Answers3