Page 1 of 1
[SOLVED] Problem in verify user password..
Posted: Mon May 24, 2004 10:09 pm
by crypdude
This is my code
Code: Select all
<?php
$query = "SELECT * FROM register_user where user_id= '$id' AND password='$pwd'";
$result = mysql_query($query);
$query_data = mysql_fetch_row($result);
IF (!$query_data[0]) {
echo("<center><p> </p><p> </p><p> </p><p> </p><p> </p>");
$error = "You have submited an incorrect login and password combination. Please try again!";
echo($error);
echo(" <a href="index.php">login user</a></center>");
}
ELSE{
echo($query);
$user= $query_data["user_type"];
$user_id= $query_data["user_id"];
echo("$user_id");
if ($user == "Data Entry")
{
echo("<br>U are ");
echo("$user!");
}
else
{
echo("<br>U are ");
echo("suppose to be not data entry!");
echo("U are ");
echo("$user!");
?>
Actually, there are many users will use my system, then when i select it by using their id and password, then system will know the type of user who currently try to access my system. I try to do my best, but not working. Can anyone help me?
Posted: Mon May 24, 2004 10:20 pm
by John Cartwright
Code: Select all
<?php
$result = @mysql_query("SELECT * FROM register_user where user_id= '$id' AND password='$pwd'");
$numrow = mysq_num_rows($result);
if ($numrows > 0){
// succesful login
if ($user == "Data Entry")
{
echo "<br>You are ".$user."!";
}
else
{
echo "<br>You are not Data Entry, you are ".$user;
}
}else{
// unsuccesful login
echo "You have submited an incorrect login and password combination. Please try again!";
echo "<center><a href='index.php'>login user</a></center>";
}
?>
This should work and is alot more clean.
?>
Posted: Mon May 24, 2004 10:47 pm
by Joe
Yup go with Phenoms method. Can i note
$result = @mysql_query("SELECT * FROM register_user where user_id= '$id' AND password='$pwd'");
$numrow = mysq_num_rows($result);
SHOULD BE:
$result = @mysql_query("SELECT * FROM register_user where user_id= '$id' AND password='$pwd'") or die("Query Error!");
$numrow = mysq_num_rows($result);
always best to check errors in the scripts and querys, lol
Joe

Posted: Mon May 24, 2004 11:04 pm
by John Cartwright
Silly me, how did I miss that... you should always have an escape to report your errors.. sometimes they don't all get shown ( unless you got error_reporting(E_ALL) on.
Posted: Mon May 24, 2004 11:45 pm
by crypdude
Thanks for all favours!
But when i trying to do so, I can't access it anymore.. It seem like can't get $numrow..even do i change it from
Code: Select all
<?php
$numrow = mysq_num_rows($result)
?>
to
Code: Select all
<?php
$numrow = mysql_num_rows($result)
?>
..
That is not my currently actual problem, my problem is how to get data from ($user) and how to check it either it is Data Entry or not? How to get it($user) from the sql query? Pls help me..!!!
May God bless!

Posted: Tue May 25, 2004 3:38 pm
by John Cartwright
K here is what I got from that: I still don't know what user is. Is it the users name? or is it they level level? what? I ask this because I have no idea what $dataentry is supposed to mean. And thirdly I'm assuming you are pulling dataentry from the database.
Code: Select all
<?php
<?php
$result = @mysql_query("SELECT * FROM register_user where user_id= '$id' AND password='$pwd'");
$numrow = @mysq_num_rows($result);
if ($numrows > 0){
$row = @mysql_fetch_array($result))
$user= $row["user"];
// succesful login
if ($user == "Data Entry")
{
echo "<br>You are ".$user."!";
}
else
{
echo "<br>You are not Data Entry, you are ".$user;
}
}else{
// unsuccesful login
echo "You have submited an incorrect login and password combination. Please try again!";
echo "<center><a href='index.php'>login user</a></center>";
}
?>
Posted: Tue May 25, 2004 4:31 pm
by Joe
Uh oh! Phenom fogot to put the error handles in again! HEHE!!!
Posted: Tue May 25, 2004 9:11 pm
by John Cartwright
hehe sorry i just used edit from my last post..
dont forget to add in your error reporting.
or else. <--- my php joke!!
HAHAH LAUGH
Posted: Wed Jun 09, 2004 8:45 pm
by crypdude
Thanks dudes! Now I just have a new life... like live after dead...hehe!