php MySQL snytax error
- by Jacksta
my scrip is supposed to look up contacts in a table and present thm on the screen to then be edited. however this is not this case. I am getting the error
Parse error: syntax error, unexpected $end in /home/admin/domains/domain.com.au/public_html/pick_modcontact.php on line 50 NOTE: this is the last line in this script.
<?
session_start();
if ($_SESSION[valid] != "yes") {
header( "Location: contact_menu.php");
exit;
}
$db_name = "testDB";
$table_name = "my_contacts";
$connection = @mysql_connect("localhost", "user", "pass") or die(mysql_error());
$db = @mysql_select_db($db_name, $connection) or die(mysql_error());
$sql = "SELECT id, f_name, l_name FROM $table_name ORDER BY f_name";
$result = @mysql_query($sql, $connection) or die(mysql_error());
$num = @mysql_num_rows($result);
if ($num < 1) {
$display_block = "<p><em>Sorry No Results!</em></p>";
} else {
while ($row = mysql_fetch_array($result)) {
$id = $row['id'];
$f_name = $row['f_name'];
$l_name = $row['l_name'];
$option_block .= "<option value\"$id\">$l_name, $f_name</option>";
}
$display_block = "<form method=\"POST\" action=\"show_modcontact.php\">
<p><strong>Contact:</strong>
<select name=\"id\">$option_block</select>
<input type=\"submit\" name=\"submit\" value=\"Select This Contact\"></p>
</form>";
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Modify A Contact</title>
</head>
<body>
<h1>My Contact Management System</h1>
<h2><em>Modify a Contact</em></h2>
<p>Select a contact from the list below, to modify the contact's record.</p>
<? echo "$display_block"; ?>
<br>
<p><a href="contact_menu.php">Return to Main Menu</a></p>
</body>
</html>
