Is this HTML in a print or echo statement?  If not, then the variables will not 
display.  You can
View Source to verify this.
----- Original Message -----
From: "Steve Jackson" <[EMAIL PROTECTED]>
To: "PHP General" <[EMAIL PROTECTED]>
Sent: Thursday, March 06, 2003 8:27 AM
Subject: [PHP] Problem updating


Been on this a couple of hours... anyone see what I'm doing wrong?
I get a result of 1 when I echo $result but it doesn't want to update at
all.
The action of this form index.php?action=update-account is just a switch
case
Which asks you to use the function update_subscriber_account() below...

<form method='post' action='index.php?action=update-account'>
   <input type='hidden' name='old_id' value='$email'>
   <tr>
     <th colspan = 2 bgcolor = '#5B69A6'>
        $title
     </th>
   </tr>
   <tr>
     <td>Real Name:</td>
     <td><input type = text name='new_realname' maxlength = 100
          value ='$realname'></td>
   </tr>
   <tr>
     <td>Preferred Name:</td>
     <td><input type = text name='new_nickname' maxlength = 100
          value ='$nickname'></td>
   </tr>
   <tr>
     <td>Company:</td>
     <td><input type = text name='new_company' maxlength = 100
          value ='$company'></td>
   </tr>
   <tr>
     <td>Email Address:</td>
     <td><input type = text name='new_email' maxlength = 100
          value ='$email'></td>
   </tr>
   <tr>
     <td>Requested Email Format:</td>";
     echo "<td><select name='new_mimetype'><option";
       if ($mimetype == 'T')
          echo " selected";
       echo ">Text Only<option";
       if ($mimetype == 'H')
          echo " selected";
       echo ">HTML</select></td>";
    print "</tr>
<tr>
   <td colspan=2 align=center>";
    display_form_button('save-changes');
    print "</td></tr></form></table></center><br>";
}
}

function update_subscriber_account()
{

db_connect();
$query = "update subscribers
set email = '$new_email',
nickname = '$new_nickname',
fullname = '$new_realname',
company = '$new_company',
mimetype = '$new_mimetype'
where email = '$old_id'";
$result = mysql_query($query)or die("Error: Could not update
query<BR>$query<BR>".mysql_error());
if (!$result)
{
echo "An error occurred when trying to update the DB";
}
else
{
echo "$result & Successfully updated the details";
}
}


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