Php Select*不显示Mysql中的记录

我写了一个问题。它没有显示来自Mysql的记录。我正在尝试从Mysql调用数据，以便更新它。但它并不是在获取数据。同样的代码在其他表单上工作，并且并没有显示标题。我不知道怎么了，但我已经尝试了所有的事情来证明。 甚至尝试直接调用数据库连接文件，但仍然没有成功

我的PHP代码

    <?php
if(isset($_GET['Competition_ID'])){

   $SelectAllPackages = mysqli_query($con, "SELECT * FROM competitions WHERE id='".$_GET['Competition_ID']."'");
   while($row=mysqli_fetch_assoc($SelectAllPackages)){

    $competition_title      = $_POST['competition_title'];
    $vimeo_url              = $_POST['vimeo_url'];
    $start_date             = $_POST['start_date'];
    $end_date               = $_POST['end_date'];
    $tickets                = $_POST['tickets'];
    $ticket_price           = $_POST['ticket_price'];
    $status                 = $_POST['status'];
   }
}

if(isset($_POST['update_competition'])){


    $up_competition_title   = $_POST['competition_title'];
    $up_vimeo_url           = $_POST['vimeo_url'];
    $up_start_date          = $_POST['start_date'];
    $up_end_date            = $_POST['end_date'];
    $up_tickets             = $_POST['tickets'];
    $up_ticket_price        = $_POST['ticket_price'];
    $up_competition_status  = $_POST['status'];

    $insertQuery = mysqli_query($con,"UPDATE competitions SET

    title                   = '$up_competition_title',
    vimeo_url               = '$up_vimeo_url',
    start_date              = '$up_start_date',
    end_date                = '$up_end_date',
    tickets                 = '$up_tickets',
    ticket_price            = '$up_ticket_price',
    status                  = '$up_competition_status'

    WHERE id='".$_GET['Competition_ID']."'
     "); 

    if($insertQuery){
        echo '<script>window.location.href = "all_competitions.php";</script>';
        }       
}


?>

我的HTML代码在这里

<div class="col-md-6">
                                            <div class="form-group">
                                                <label class="control-label">Competition Title</label>
                                                <input type="text" id="firstName" name="competition_title" class="form-control" value="<?php echo $title; ?>">
                                                </div>
                                        </div>

你为什么不使用

if (isset($_POST['Competition_ID'])) {
  $id = mysqli_real_escape_string($conn, $_POST['Competition_ID']);
{
 if (empty($id)) {
    array_push($errors, "ID is required");
  }
if (count($errors) == 0) {
    $query = "SELECT * FROM competition WHERE Competition_ID='$id'";
    $results = mysqli_query($conn, $query);
}

---

您是否尝试直接使用phpmyadmin回显查询？在$\u GET['Competition\u ID']中存在SQL注入，请使用“对不起”命令，通过直接使用phpmyadmin回显查询而未获得您，请选择SQL查询选择*。复制查询将使用phpmyadmin执行。现在您可以在SQL查询中获得任何错误。请提供帮助，我哪里错了？谢谢。我将负责SQL注入，但我仍在努力进行数据显示。