Question

我有这个表单并且在这个表单之外，我可以从带有帖子的单选按钮获得2个值。但是，我希望能够发送另一个值winch在我的while循环中$fieldname变量与我的表单。我只是不知道该怎么做。

这是我的代码：

$result = mysqli_query($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>";
  }

这是我得到帖子的代码。并更新我的数据库

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);
}}

Answer 1

顺便说一下thnx家伙...输入隐藏使其工作

size_t median = sizeArray / 2;

return (pArray[median] + pArray[median+1]) / 2;

使用PHP发布多个表单值

1 个答案: