我正在運行while循環並從數據庫中獲取3條記錄。然後在同一頁面上更新它。每個記錄都有提交按鈕。但編輯完成後,當我提交表單時,它只捕獲最後一條記錄的值,並用最後一條記錄值更新其他行。如果有人幫助我,我會非常感激。記住它捕獲確切的(id),但其他參數只是最後一行。PHP表格更新更新形式
<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");
}
感謝。 –
@ZainAbid歡迎:) –