我不明白，我在这段代码中没有看到任何错误，但是有这个错误，请帮助：
mysql_fetch_array（）期望参数1是资源问题

<?php

      $con = mysql_connect("localhost","root","nitoryolai123$%^");
    if (!$con)
      {
      die('Could not connect: ' . mysql_error());
      }

    mysql_select_db("school", $con);
       $result = mysql_query("SELECT * FROM student WHERE IDNO=".$_GET['id']);
    ?>


                           <?php while ($row = mysql_fetch_array($result)) { ?>             
                                     <table class="a"  border="0" align="center" cellpadding="0" cellspacing="1" bgcolor="#D3D3D3">
    <tr>

    <form name="formcheck" method="get" action="updateact.php" onsubmit="return formCheck(this);">
    <td>
    <table  border="0" cellpadding="3" cellspacing="1" bgcolor="">
    <tr>

    <td  colspan="16" height="25"  style="background:#5C915C; color:white; border:white 1px solid; text-align: left"><strong><font size="2">Update Students</td>


    <tr>
    <td width="30" height="35"><font size="2">*I D Number:</td>
    <td width="30"><input  name="idnum" onkeypress="return isNumberKey(event)" type="text" maxlength="5" id='numbers'/ value="<?php echo $_GET['id']; ?>"></td>
    </tr>

    <tr>
    <td width="30" height="35"><font size="2">*Year:</td>
    <td width="30"><input  name="yr" onkeypress="return isNumberKey(event)" type="text" maxlength="5" id='numbers'/ value="<?php echo $row["YEAR"]; ?>"></td>

<?php } ?>

我只是试图在表单中加载数据，但不知道为什么会出现该错误。这里可能有什么错误？