Quote:
Originally Posted by dirtysouth
Got a page where users can edit their account details ie: useredit.php
That page posts to "do-useredit.php
User must be logged in. Here's some code:
useredit.php
PHP Code:
<?php
require_once('../mysql_connect.php');
session_start();
if(isset($_SESSION['id']))
{
$mn = trim($_SESSION['id']);
require_once('../mysql_connect.php');
$query = "SELECT * FROM shoppingmembers WHERE member_name = '$mn'";
$result = mysql_query($query);
$row = mysql_fetch_array($result, MYSQL_ASSOC);
}
echo'
<form name="form1" method="post" action="do-useredit.php">
<input type="text" name="bill_fname" value="' . $row['bill_fname'] . '" style="font-size:9px"><br />
<input type="submit" name="submit" value="Update" />';
?>
Here's the second page:
do-useredit.php
PHP Code:
<?php
require_once('../mysql_connect.php');
session_start();
$_GET['mn'];
if(isset($_SESSION['id']))
{
$mn = trim($_SESSION['id']);
require_once('../mysql_connect.php');
$query = "SELECT * FROM shoppingmembers WHERE member_name = '$mn'";
$result = mysql_query($query);
$row = mysql_fetch_array($result, MYSQL_ASSOC);
}
mysql_query("UPDATE shoppingmembers SET bill_fname = '$bill_fname' WHERE member_name = 'admin'")or die(mysql_error());
echo "Record Updated";
print $mn;
?>
Problem: When I have session_start in the top of do-useredit.php, it doesn't UPDATE. When I comment it out it works fine. The trouble is, I can't pass the member_name into the query, hence in the above I simply force it to UPDATE WHERE member_name = 'admin' (me). Any ideas? I have paypal funds avail. TIA! |
You're running this segment of code:
PHP Code:
if(isset($_SESSION['id']))
{
$mn = trim($_SESSION['id']);
require_once('../mysql_connect.php');
$query = "SELECT * FROM shoppingmembers WHERE member_name = '$mn'";
$result = mysql_query($query);
$row = mysql_fetch_array($result, MYSQL_ASSOC);
}
before the update SQL below, so I'd guess your problem has something to do with trying to dray the session ID out and put it into the SQL query. Try changing:
PHP Code:
$result = mysql_query($query);
to:
PHP Code:
$result = mysql_query($query) or die("Error: ".mysql_error());
and see if that gives you an SQL error.