将'SELECT'值发布到下一页 PHP

Question

这是表单代码：

<?php
$con=mysqli_connect("localhost","user","pass","db");
// Check connection
if (mysqli_connect_errno())
  {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
  }
$result = mysqli_query($con,"SELECT ID, NAME FROM b_sonet_group ORDER BY ID DESC");

echo "<form class='form-vertical login-form' action='step-2.php' method='POST'>";
echo "<h4 class='form-title'>Step One: Choose Your Project</h4><div class='control-group'><div class='controls'>";              
echo "<select>";
echo "<option value=''>Choose Your Project</option>";
while($row = mysqli_fetch_array($result))
  {
  echo "<option name='ID' value='" . $row['ID'] . "'>" . $row['NAME'] . "</option>";
  }
  echo "</select>";
  echo "</div></div>";
  echo "<div class='form-actions'><button type='submit' name='submit' class='btn green pull-right'>Proceed to Step Two <i class='m-icon-swapright m-icon-white'></i></button></div></form>";

mysqli_close($con);
?>

我需要在第 2 页上放置什么，以便从上一页的表单中检索值 ID，以及如何打印它以便我可以检查它是否正确？

我知道很简单，但我的大脑已经收拾好回家了。

Answer 1

您需要将

select更改为

echo "<select name="project">";

在第二页上，您可以通过以下方式获取值

echo $_POST['project'];

Answer 2

您需要将名称从选项移动到您的选择。然后回显 $_POST['id'];