2015-11-02 36 views
0

我有這種形式,出於這種形式,我可以從郵政單選按鈕中獲得2個值。但是,我想能夠發送另一個價值絞車在我的while循環$fieldname變量與我的形式。我只是不知道該怎麼做。用PHP發佈多個表單值

這是我的代碼:

$result = mysqli_query($con,"SELECT * FROM Velden"); 
while($row = mysqli_fetch_array($result)) { 
echo "<div>"; 
echo "<h1>".$row['name']."</h1>"; 
echo "<h3>".$row['locatie']."</h3>"; 
echo '<img src="images/'.$row['photo'].'" width="120px" height="120px"/>'; 
echo "<p>".$row['aanwezig']."</p>"; 
$namefield = $row['name']; 
$players = mysqli_query($con, "SELECT name, user_status FROM veld_user WHERE  user_status=1 AND name='$namefield'"); 
echo "veld: ".$row['name']."<br />"; 
$number = mysqli_num_rows($players); 
echo "Aantal spelers aanwezig: ".$number."<br /><br />";  
?> 
<form action="" method="post" id="registerForm"> 
<table class="form imageFrom"> 
     <tr> 
      <td><input checked type="radio" name="status" value="1"/> aanwezig</td> <?php if (isset($_POST['status']) && $_POST['status']=='1') echo ' STATUS="aanwezig"';?> 
      <td><input checked type="radio" name="status" value="0"/> afwezig</td><?php if (isset($_POST['status']) && $_POST['status']=='0') echo ' STATUS="afwezig"';?> 
     </tr> 
     <tr> 
      <td><input type="submit" name="submit" value="submit" class="knop"/></td> 
     </tr> 
    </table> 
</form><?php 
    echo"</div>"; 
    } 

這是代碼中我得到這個職位。並更新我的數據庫

if(isset($_POST['submit'])){ 
if (isset($_POST['status']) && $_POST['status']=='1'){ 
$sql = "UPDATE veld_user SET user_status = 1 WHERE id=".$user->data()->id;} 
elseif (isset($_POST['status']) && $_POST['status']=='0'){ 
$sql = "UPDATE veld_user SET user_status = 0 WHERE id= ".$user->data()->id;} 
    if (mysqli_query($con, $sql)) { 
    Session::flash('home', 'update success'); 
} else { 
    echo "Error updating record: " . mysqli_error($con); 
}} 
+0

'' – AbraCadaver

+2

你很容易[sql注入攻擊](http://bobby-tables.com)。 –

+1

@MarcB他們並不關心我確定,但我們仍然嘗試:-( – AbraCadaver

回答

0

順便說一句日Thnx傢伙...輸入隱藏着使工作

<input type="hidden" name="fieldname" value="<?php echo $namefield?>" />