问题：如何解决我在处理会话时遇到的问题，因为它返回了不正确的值。情况：我在表中的这个会话变量上有问题。我使用 while 循环将数据从数据库添加到表中。这是我的代码：
<form action="edit2.php" method="get">
<?php
$link = mysql_connect("localhost", "root", "root");
mysql_select_db("ispot", $link);
$result4 = mysql_query("SELECT * FROM user_ispot", $link);
$num_rows = mysql_num_rows($result4);
$result = mysqli_query($con,"SELECT * FROM complaints");
echo "<table border='1'>
<tr>
<th>Id Number</th>
<th>Category</th>
<th>Problem</th>
<th>Date Reported</th>
<th>Complaint ID </th>
<th>Action</th>
</tr>";
while($row = mysqli_fetch_array($result))
{
echo "<tr>";
echo "<td name=id_num>" . $row['id_number'] . "</td>";
$_SESSION['favcolor'] = "$row[id_number]";
echo "<td name=remarks>" . $row['remarks'] . "</td>";
echo "<td name=status>" . $row['status'] . "</td>";
echo "<td name=date>" . $row['date_reported'] . "</td>";
echo "<td>" . "<INPUT TYPE = text Name = cid VALUE = " . $row['complaint_id'] . ">" . "</td>";
echo "<td>" . "<INPUT TYPE = Submit Name = Submit1 VALUE =Edit>" . "</td>";
echo "</tr>";
}
echo "</table>" ;?>
它看起来像这样：
如您所见，有一个编辑按钮，我可以在其中编辑表格中的特定行。当我按下编辑按钮时，这将显示：
请注意，用户 ID 错误，我该如何解决？因为这里发布的用户 ID 是插入表中的最后一个 user_id。
这是我的第二张图片的代码：
<b>Date:</b> <input type='text' name='today' placeholder='<?php echo $today ?>' disabled='disabled'> <br><br>
<b>User ID:</b> <input type='text' disabled='disables' name='userid' placeholder='<?php
//$comid = $_GET["cid"];
//echo $userid;
echo $_SESSION['userid'];
//$result = mysqli_query($con,"SELECT * FROM complaints WHERE id = XXX");
//$row = mysqli_fetch_assoc($result);
//print_r($row);
//$result2 = mysql_query("SELECT * FROM complaints WHERE complaint_id = '$comid'", $link);
//$result2 = mysql_query("SELECT * FROM complaints", $link);
//while($row = mysql_fetch_assoc($result2))
//{
//echo $row['id_number'];
//}
?>'></br><br>
任何帮助，将不胜感激。谢谢你。