Updating Member Profile()

[tobimichigan]

Hi Code Gurus,

<?php
include("cn.php");

$pfno=$_GET['pfno'];
$result=mysql_query("SELECT (*) FROM user_table WHERE pfno=$pfno");
//$result = mysql_fetch_assoc($select);
$num=mysql_num_rows($result);
//mysql_close();

$i=0;
while ($i < $num) {
$amountd=mysql_result($result,$i,"amountd");
$department=mysql_result($result,$i,"department");
$email=mysql_result($result,$i,"email");
$fname=mysql_result($result,$i,"fname");
$oname=mysql_result($result,$i,"oname");
$lname=mysql_result($result,$i,"lname");
$lga=mysql_result($result,$i,"lga");
$marital=mysql_result($result,$i,"marital");
$Nationalty=mysql_result($result,$i,"Nationalty");
$pfno=mysql_result($result,$i,"pfno");
$residentialadd=mysql_result($result,$i,"residentialadd");
$sex=mysql_result($result,$i,"sex");
$soorigin=mysql_result($result,$i,"soorigin");
$telno=mysql_result($result,$i,"telno");
//Space For Code

$query = ("UPDATE user_table SET amountd = '$amountd', department = '$department', email = '$email', fname = '$fname', lga = '$lga', lname = '$lname',marital='$marital', Nationalty='$Nationalty',oname='$oname', residentialadd='$residentialadd', soorigin='$soorigin',telno='$telno', WHERE pfno = '$pfno'");
mysql_query($query);
echo //"Record Updated";
mysql_close();

++$i;
}

$amountd=$_POST['amountd'];
$department=$_POST['department'];
$email=$_POST['email'];
$fname=$_POST['fname'];
$lga=$_POST['lga'];
$lname=$_POST['lname'];
$marital=$_POST['marital'];
$Nationalty=$_POST['Nationalty'];
$oname=$_POST['oname'];
$residentialadd=$_POST['residentialadd'];
$soorigin=$_POST['soorigin'];
$telno=$_POST['telno'];

?>

 

There are 2 things I'd love to do.

Firstly, display the existing member's profile data on an html form.

Secondly, update the previous values stored  by new ones with sql update as above.

 

When I preview in an explorer on the form it gives me this-><? echo $lname; ?> as a form return value instead of showing the current member information and after clicking submit, updating the profile accordingly.

This error:

"Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in line 24" is also coming out of the browser top page. Where line 24 =

$num=mysql_num_rows($result);

Please gurus help me out...

[kartul]

if i get this right then to display current db values in input fields:

$sql = "select * from users where id=$id";
$row = mysql_fetch_assoc($sql);
//now the form
echo "<input type="text" name="location" value="\".$row['location'].\"">"

[MasterACE14]

$sql = "select * from users where id=$id";
$row = mysql_fetch_assoc($sql);
//now the form
echo "<input type=\"text\" name=\"location\" value=\"".$row['location']."\" />"; // missing a few slashes and a semi colon 

[tobimichigan]

$sql = "select * from users where id=$id";
$row = mysql_fetch_assoc($sql);
//now the form
echo "<input type=\"text\" name=\"location\" value=\"".$row['location']."\" />"; // missing a few slashes and a semi colon 

 

How is " <input type=\"text\" name=\"location\" value=\"".$row['location']."\" />"; " supposed to be encapsed with

<?php ?>

tags? in the form?

[tobimichigan]

Please could someone tell how  " <input type=\"text\" name=\"location\" value=\"".$row['location']."\" />"; " supposed to be encapsed in code tags?