表单不发布变量/mySQL 查询不正确搜索

我正在尝试为足球网站的管理部分创建一个球员编辑系统。该过程如下：

"教练.php"，他们就可以通过下拉菜单选择他们想要查看的教练课程，然后填充"玩家"下拉菜单(通过下面的js完成(

教练.php表单

            <form id="form1" name="form1" method="post" action="coachplayer.php?id=' .$id. '">
            <label>Activity :</label>
            <select name="activity" class="activity">
                <option selected="selected">--Select Activity Group--</option>
                    <?php
                    include('dbconnect.php');
                    $sql=mysql_query("select activity from coaches where username='$coach'");
                        while($row=mysql_fetch_array($sql))
                      {
                          $activity2=explode(",",$row["activity"]);
                          foreach ($activity2 as $activity)
                         echo '<option value="'.$activity.'">'.$activity.'</option>';
                     } ?>
                </select> <br/><br/>
                <label>Player :</label> <select name="username" class="username">
                <option selected="selected">--Select Player--</option>
            </select>
            <input name="thisID" type="hidden" value="<?php echo $id; ?>" />
            <input type="submit" name="button" id="button" value="Log In" />
        </form>

教练.php js 函数

<script type="text/javascript">
$(document).ready(function()
{
$(".activity").change(function()
{
var activity=$(this).val();
var dataString = 'activity='+ activity;
$.ajax
({
type: "POST",
url: "ajax_city.php",
data: dataString,
cache: false,
success: function(html)
{
$(".username").html(html);
} 
});
});
});
</script>
<style>
label
{
font-weight:bold;
padding:10px;
}
</style>

如上面的js所示，玩家列表通过一个单独的页面处理，其中包含一个查询，如下所示：

<?php
if($_POST['activity'])
{
$activity=$_POST['activity'];
$sql=mysql_query("SELECT id, username FROM stats WHERE activity='$activity'");
while($row=mysql_fetch_array($sql))
{
$id=$row['id'];
$username=$row['username'];
$activity=$row['activity'];
echo '<option value="'.$username.'">'.$username.'</option>';
}
}
?>

完成所有这些操作后，教练提交表格，将他们带到教练.php。这就是问题开始的地方。

CoachPlayer.php是一个模板页面，空白字段填充了Echo，以便在必要时回显玩家详细信息。运行查询以获取所选玩家的 ID，显示他们的详细信息并填充页面。但是，相反，如果查询无法通过$playerCount找到匹配的结果，则通常会出现如下所示，例如"Player 不存在"。

教练员.php SQL 查询

<?php 
// Check to see the URL variable is set and that it exists in the database
if (isset($_GET['id'])) {
    // Connect to the MySQL database  
    $targetU = preg_replace('#[^0-9]#i', '', $_GET['id']); 
    // Use this var to check to see if this ID exists, if yes then get the player 
    // details, if no then exit this script and give message why
    $sql = mysql_query("SELECT * FROM stats WHERE id='$targetU' LIMIT 1");
    $playerCount = mysql_num_rows($sql); // count the output amount
    if ($playerCount > 0) {
        // get all the product details
        while($row = mysql_fetch_array($sql)){ 
             $id = $row["id"];
             $username = $row["username"];
             $position = $row["position"];
             $activity = $row["activity"];
             $agegroup = $row["agegroup"];
             $coach = $row["coach"];
             $goals = $row["goals"];
             $assists = $row["assists"];
             $cleans = $row["cleans"];
             $motm = $row["motm"];
             $attend = $row["attend"];
         }
    } else {
        echo "Player doesn't exist.";
        exit();
    }
} else {
    echo "Data to render this page is missing.";
    exit();
}
?>

我相信你可以说，我不太擅长编码员，所以很可能这是一个需要改变的简单变量，但我出错的任何想法都将不胜感激。

提前谢谢你。