[SOLVED] Resource id
Posted: Wed Nov 24, 2004 11:34 am
Hi, I am making a site for my school and one page that I am making is an "Update your profile" page.
Here is my code:
When I echo the $subject, $phone, $email, and $biography in text fields, it instead echos "Resource id #" then a number. What is the error? Thanks a ton!
Here is my code:
Code: Select all
$username = "usernmae";
$password = "password";
$database = "db";
mysql_connect("localhost","$username","$password") or die ("Unable to connect to MySQL server.");
$db = mysql_select_db("$database") or die ("Unable to select requested database.");
$subjects = mysql_query("SELECT subjects FROM teachers WHERE id='$_COOKIE[id]'");
$phone = mysql_query("SELECT phone FROM teachers WHERE id='$_COOKIE[id]'");
$email = mysql_query("SELECT email FROM teachers WHERE id='$_COOKIE[id]'");
$biography = mysql_query("SELECT biography FROM teachers WHERE id='$_COOKIE[id]'");
IF($_POST[submit]){
$encryptoldpassword = md5($_POST[oldpassword]);
$encryptnewpassword1 = md5($_POST[newpassword1]);
$encryptnewpassword2 = md5($_POST[newpassword2]);
IF($_POST[newpassword1] == "" AND $_POST[newpassword2] == "" AND $_POST[oldpassword] == ""){
$sql = "INSERT INTO teachers (subjects,phone,email,biography) VALUES ('$_POST[subjects]','$_POST[phone]','$_POST[email]','$_POST[biography]') WHERE id='$_COOKIE[id]'";
$result = mysql_query($sql);
}ELSEIF($encryptnewpassword1 == "$encryptnewpassword2" AND $encryptoldpassword == "$_COOKIE[password]"){
$sql = "INSERT INTO teachers (password,subjects,phone,email,biography) VALUES ('$encryptnewpassword2','$_POST[subjects]','$_POST[phone]','$_POST[email]','$_POST[biography]') WHERE id='$_COOKIE[id]'";
$result = mysql_query($sql);
}}