[SOLVED] Need Guru Help! PHP Login Not Working!!!

coldfiretech · August 20, 2008

Hello!

I am trying to make a login system to will login to differnt pages depending on the subscription type

There are 3 different subscription types single, partner, family.

I made the $subtype varible because i couldnt figure out if it was even getting that data

So if i go Print $subtype it shows the data in there but i for some reason i cant get the nested elseif statment to work.. If someone could please show me what i am doing wrong i would greatly appreciate it.!! Thanks.

<?php
$db_host = 'localhost';
$db_user = '****';
$db_pass = '*****';
$db_db = '******';
$con = mysql_connect($db_host, $db_user, $db_pass) or die('MySQl Connection Error:'.mysql_error());
mysql_select_db($db_db, $con) or die('MySQL Error: Cannot select table');
$user = $_POST['login'];
$passwd = $_POST['password'];
$sql0 = "SELECT * FROM users";
$results = mysql_query($sql0,$con);
while($row = mysql_fetch_array($results))
if ($user == $row['login'] && $passwd == $row['password']) {
$subtype = $row['subtype'];

if ($subtype="atype") {
print "<META HTTP-EQUIV=\"refresh\" content=\"2; URL=/*****/a.php\">";
} elseif ($subtype="btype") {
print "<META HTTP-EQUIV=\"refresh\" content=\"2; URL=/******r/b.php\">";
}elseif ($subtype="ctype") {
print "<META HTTP-EQUIV=\"refresh\" content=\"2; URL=/******/c.php\">";
}else {

}

} else {
print "<META HTTP-EQUIV=\"refresh\" content=\"2; URL=/index.html\">";
}
mysql_close ($con);
?>

Third_Degree · August 20, 2008

I haven't read all your code but try changing:

//This:
while($row = mysql_fetch_array($results)) {

//To This:
while($row = mysql_fetch_assoc($results)) {

coldfiretech · August 20, 2008

I dont think that is the problem becuase i can read the field. I think my if statement is wrong because i can get into the a type page.....

cooldude832 · August 20, 2008

oh the operators (this isn't a guru question )

the operators using the equal sign

= = Set something equal to that value

== = Compare the two values "weak"

=== = Bitwise comparasion "strong"

!= = Not equal to value before

See the problem

Third_Degree · August 20, 2008

ah

$subtype == "atype"

not

$subtype="atype"

sorry cooldude didn't see your post. yeah...

cooldude832 · August 20, 2008

Third degree do you get this Red warning that pops up if you try and respond (such as u just did) after someone else posted?

I've heard ppl don't' get it and I wanted to know if its a myth or not.

akitchin · August 20, 2008

== = Compare the two values "weak"

=== = Bitwise comparasion "strong"

that's not a bitwise comparison. it is an identical comparison - it checks that both the value and the type of the variables match. bitwise operators are far different than that.

"weak" and "strong" are not really accurate adjectives. == is adequate in almost all cases, except when you're comparing values that may evaluate to equivalent because they're converted to integers (for example: FALSE and integer 0, if you're comparing strpos() returns).

the red warning is not a myth. people just may not have read the post before going to reply (therefore don't get the warning but still haven't read it).

cooldude832 · August 20, 2008

miswrote that meant to say type equal not just value

coldfiretech · August 20, 2008

I dont get it.... ???

Tried fixing the == Still nothing.. Can one of you please try this code out. I really think it is the if statement because the atype part of it works just not the b and c types !

Thanks

<?php
$db_host = 'localhost';
$db_user = '****';
$db_pass = '*****';
$db_db = '******';
$con = mysql_connect($db_host, $db_user, $db_pass) or die('MySQl Connection Error:'.mysql_error());
mysql_select_db($db_db, $con) or die('MySQL Error: Cannot select table');
$user = $_POST['login'];
$passwd = $_POST['password'];
$sql0 = "SELECT * FROM users";
$results = mysql_query($sql0,$con);
while($row = mysql_fetch_array($results))
if ($user == $row['login'] && $passwd == $row['password']) {
$subtype = $row['subtype'];

if ($subtype=="atype") {
print "<META HTTP-EQUIV=\"refresh\" content=\"2; URL=/*****/a.php\">";
} elseif ($subtype=="btype") {
print "<META HTTP-EQUIV=\"refresh\" content=\"2; URL=/******r/b.php\">";
}elseif ($subtype=="ctype") {
print "<META HTTP-EQUIV=\"refresh\" content=\"2; URL=/******/c.php\">";
}else {

}

} else {
print "<META HTTP-EQUIV=\"refresh\" content=\"2; URL=/index.html\">";
}
mysql_close ($con);
?>

cooldude832 · August 20, 2008

Tried fixing the == Still nothing.. Can one of you please try this code out.

We don't have your database or dump so we can't try it

add in

echo "<br /><br />".$subtype."<br /><br />";

right before your if statements to see if the value is reasonable

coldfiretech · August 20, 2008

I tried echoing the $subtype varible and it prints the data accordingly..

Third_Degree · August 20, 2008

Third degree do you get this Red warning that pops up if you try and respond (such as u just did) after someone else posted?

I've heard ppl don't' get it and I wanted to know if its a myth or not.

I used to get it, but I turned it off in profile settings.

cooldude832 · August 20, 2008

and it doesn't output any of your three meta refreshes?

see if its going right to your final if statement

also case must MaTcH

At third degree: ppl shouldn't be able to turn it off its very helpful

coldfiretech · August 20, 2008

Its not making it past the first one!!!

Sorry if i seem a little grumpy i have been working on this for hours and its getting to me!

Grr!

cooldude832 · August 20, 2008

give me the exact output

also try using a better looking tab structure

<?php
$db_host = 'localhost';
$db_user = '****';
$db_pass = '*****';
$db_db = '******';
$con = mysql_connect($db_host, $db_user, $db_pass) or die('MySQl Connection Error:'.mysql_error());
mysql_select_db($db_db, $con) or die('MySQL Error: Cannot select table');
$user = $_POST['login'];
$passwd = $_POST['password'];
$sql0 = "SELECT * FROM users";
$results = mysql_query($sql0,$con);
while($row = mysql_fetch_array($results))
if ($user == $row['login'] && $passwd == $row['password']) {
$subtype = $row['subtype'];
if ($subtype=="atype") {
	print "<META HTTP-EQUIV=\"refresh\" content=\"2; URL=/*****/a.php\">";
}
elseif ($subtype=="btype") {
	print "<META HTTP-EQUIV=\"refresh\" content=\"2; URL=/******r/b.php\">";
}
elseif ($subtype=="ctype") {
	print "<META HTTP-EQUIV=\"refresh\" content=\"2; URL=/******/c.php\">";
}	
else {

}

} 
else {
print "<META HTTP-EQUIV=\"refresh\" content=\"2; URL=/index.html\">";
}
mysql_close ($con);
?>

cooldude832 · August 20, 2008

a more pressing problem is in the fact that your login structure is poorly written

change

$sql0 = "SELECT * FROM users";

to

$sql0 = "SELECT * FROM users Where login = '".mysql_real_escape_string($_POST['login'])."' and password = '".mysql_real_escape_string($_POST['password'])."'";

and then instead of while($row = fetch put

if(mysql_num_rows($result) >0){


That is how a login should work

coldfiretech · August 20, 2008

Okay.

Exact output is its not working! Okay. Again, its not getting passed the first if statement. If i echo $subtype and try logging in with the 3 different types it prints atype for the a type and btype for the btype and ctype for the c type! Okay, now it only redirects on the a type plan. I dont know if i can say this any more clearly.

I guess ill just have to figure it out. Im not trying to stress anyone out here.

cooldude832 · August 20, 2008

I think you need to chill out and actually tell us what its doing

I don't see in your script where it can echo "not working" unless you are hiding something so a bit more info will help us help you

akitchin · August 20, 2008

if the first user that gets pulled from the database doesn't match your user's info (the POST info), it will redirect you to index.php. that's because in your else{}, you tell it to do so. either drop the else{} echoing from the while() loop, or place it after the while() loop altogether.

coldfiretech · August 20, 2008

Im fine. Im just really agrivated with this code.

Like i said in my first post. I just started writing php yesterday. So thanks for the comment on my poorly written code. Im sure you wrote your first login system flawlessly.

And no, im not hidding something

Ill post the file so you can see it for yourself.

[attachment deleted by admin]

akitchin · August 20, 2008

ignore cooldude's post and have a look at mine. i think you'll find that answers your question.

cooldude832 · August 20, 2008

I'm trying to help you by saying you are over workin the server for that

I am going to rewrite this for you because I don't want to see you discouraged from php but don't expect hand outs

<?php
$db_host = 'localhost';
$db_user = '****';
$db_pass = '****';
$db_db = '*****';
$con = mysql_connect($db_host, $db_user, $db_pass) or die('MySQl Connection Error:'.mysql_error());
mysql_select_db($db_db, $con) or die('MySQL Error: Cannot select table');

$user = mysql_real_escape_string($_POST['login'])
$password = mysql_real_escape_string($_POST['password']);

$q = "
SELECT 
	login, subtype
FROM `users`
Where
	login = '".$user."' AND 
	password = '".$password."'
";
$r = mysql_query($q,$con) or die(mysql_error()."<br /><br />".$q);
$results = mysql_query($sql0,$con);
if(mysql_num_rows($result) >0){
$row = mysql_fetch_assoc($r);

switch ($row['subtype']){

	case "single":
		die(header("location: /****/single.php"));
	break;
	case "partner":
		die(header("location: /****/partner.php"));
	break;
	case "family":
		die(header("location: /***/family.php"));
	break;
	default:
		die(header("location: index.html?msg=invalid subtype"));
}
}
else{
die(header("location: index.html?msg=invalid login"));
}
mysql_close ($con);
?>

coldfiretech · August 20, 2008

Moved the other redirect to the bottom. Still nothing

Got this error though

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in C:\wamp\www\cellsavior\userAuth.php on line 13

Login Failed

cooldude832 · August 20, 2008

sorry missed that

<?php
$db_host = 'localhost';
$db_user = '****';
$db_pass = '****';
$db_db = '*****';
$con = mysql_connect($db_host, $db_user, $db_pass) or die('MySQl Connection Error:'.mysql_error());
mysql_select_db($db_db, $con) or die('MySQL Error: Cannot select table');

$user = mysql_real_escape_string($_POST['login'])
$password = mysql_real_escape_string($_POST['password']);

$q = "
SELECT 
	login, subtype
FROM `users`
Where
	login = '".$user."' AND 
	password = '".$password."'
";
$r = mysql_query($q,$con) or die(mysql_error()."<br /><br />".$q);
if(mysql_num_rows($r) >0){
$row = mysql_fetch_assoc($r);

switch ($row['subtype']){

	case "single":
		die(header("location: /****/single.php"));
	break;
	case "partner":
		die(header("location: /****/partner.php"));
	break;
	case "family":
		die(header("location: /***/family.php"));
	break;
	default:
		die(header("location: index.html?msg=invalid subtype"));
}
}
else{
die(header("location: index.html?msg=invalid login"));
}
mysql_close ($con);
?>

coldfiretech · August 20, 2008

Thank you. I tried that code you just posted... I do appreciate your help. And i dont expect handouts.

Its highlighting this line

"$password = mysql_real_escape_string($_POST['password']);"

and when i run it it throws this

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in C:\wamp\www\cellsavior\userAuth.php on line 13

Login Failed

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[SOLVED] Need Guru Help! PHP Login Not Working!!!

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