$ _POST ['变量']不起作用

Question

我$userid从$ _POST 获得了一个问题.我以前做了很多次,所以我不确定我突然做错了什么.

提交给user_confirm.php的表单

<?php
//confirm user function
  function confirmUsers() {
    //make connection global
      global $con;

    //set user variables
    $userquery =  mysqli_query($con, "SELECT * FROM users WHERE userlevel = 0");

    //echo list
   echo '<center><form name="userConfirm" action="functions/user_confirm.php" method="post">';
     echo '<select name="confirmUser">';
       while ($row = mysqli_fetch_array($userquery)) {
         echo "<option value='" . $row['userid'] ."'>" . $row['username'] ."</option>"; 
         //in viewing element, the userid is displaying properly
       }
     echo '<input type="submit" value="Confirm User">';
     echo '</select>';
     echo '</form></center>';
  }
?>

user_confirm.php

<?php
//include db connect
  include ("db_con.php");

//set variable names
   $userid = $_POST['userid'];

//start session
  session_start();

echo $userid;


?>

如您所见,我只是试图回应从表单传递的变量.它不起作用,我完全困惑为什么,任何想法？

如果需要这里是db_con.php

<?php
$con=mysqli_connect("localhost","user","pw","db");
// Check connection
if (mysqli_connect_errno())
  {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
  }
?>

Answer 1

你没有名为userid的表单字段,也许你的意思是confirmUser字段:

$userid = $_POST['confirmUser'];