这个代码出了什么问题?它正在通过“；1〃；在表单处理中

我正在制作管理面板，这样用户就可以通过这个帮助pppp编辑页面的文本了，请。。。网页上的数据是通过这里的数据库显示的，我给了他编辑数据和更新数据的选项，这将在文本框中打开数据，当用户点击botton更新时，它将更新数据库中的数据，但问题是，当点击更新时，表单没有在文本中传递值，而是在传递值"1"

<?php
$query2 = mysqli_query($con, "SELECT * FROM home");
$row    = mysqli_fetch_array($query2);    
$pic1 = ucfirst($row['pic1']);
$pic2 = ucfirst($row['pic2']);
$pic3 = ucfirst($row['pic3']);
$pic4 = ucfirst($row['pic4']);
$pic5 = ucfirst($row['pic5']);
$pic6 = ucfirst($row['pic6']);
$pic7 = ucfirst($row['pic7']);
$pic8 = ucfirst($row['pic8']);
$par1 = ucfirst($row['par1']);
$par1pic = ucfirst($row['par1pic']);
$par2 = ucfirst($row['par2']);
$side_pic1 = ucfirst($row['side_pic1']);
$side_pic2 = ucfirst($row['side_pic2']);
?>
<form method="POST" action="update_home.php">
    Paragraph no 1:
    <textarea name="comment" rows="5" cols="40"><?php echo $par1; ?></textarea>
    First name:
    <input type="text" name="fname"><br>
    <input type="submit" name="nw_update" value="Update_1"/><br><br>
</form>
<form method="POST" action="update_home.php">
    Paragraph no 2:
    <textarea name="comment1" rows="5" cols="40"><?php echo $par2; ?></textarea>
    enter code here
    <input type="submit" name="nw_update" value="Update_2"/><br><br>
</form>

<?php
include("dbconnection.php");
$value = isset($_POST["nw_update"]);
$fname = isset($_POST["fname"]);
echo $fname;
echo $value;
if ($value == "Update_1") {
    $par = isset($_POST["comment"]);
    $query2 = mysqli_query($con, "UPDATE home SET par1='$par' where id='1'");
}
if ($value == "Update_2") {
    $par2 = isset($_POST["comment1"]);
    $query2 = mysqli_query($con, "UPDATE home SET par2='$par2' where id='1'");
}
header("location : edit_home.php");
?>