Help with my php code

[senpai91]

This is a very simple and very basic php code..
pero bisag unsaon nako di man jud mo True ang 'if' statement.
unsay possible problem ani?

this is a very simple username and password authentication
++++++++++++++++++++++++++++++++++++++++++++++++++ +

<?php
$con=mysql_connect("localhost", "root", "") or die(mysql_error());
$db=mysql_select_db("user", $con) or die(mysql_error());


$user=$_POST['user'];
$pass=$_POST['pass'];


$query=('SELECT * FROM USERS WHERE (USERNAME = "$user" and PASSWORD = "$pass")');
$result=mysql_query($query);


while($row=mysql_fetch_array($result))


{
$username = $row['USERNAME'];
$password = $row['PASSWORD'];
}
if($username==$user and $password == $pass)
{
include('home.html');
}
else
{
echo"FAILED TO LOGIN!";
}


?>

any thoughts? please help

[muchacha00]

paminaw gud og tarong TS..... basta ang g ingon ni teacher timan-e kong dili true.. false gud na siya....

[boang3000]

$result=mysql_query($query);

$count = mysql_num_rows($result);
if($count == 1){
// naka sud naka
}

[senpai91]

^ dude I did what you post but...

+++++++++++++++++++++++++++++++

<?php
$con=mysql_connect("localhost", "root", "") or die(mysql_error());
$db=mysql_select_db("user", $con) or die(mysql_error());


$user=$_POST['user'];
$pass=$_POST['pass'];


$query=('SELECT * FROM USERS WHERE USERNAME = "$user" and PASSWORD = "$pass"') or die(mysql_error());
$result=mysql_query($query);


$count = mysql_num_rows($result);
if($count == 1)
{
echo"Success";
}
else
{
echo"FAILED TO LOGIN!";
}

?>

this is still not working...

ang db ani kay naa nay sud username ug password...pero still false gi hapon ang iyang statement...

ngano kaha ni?

[boang3000]

$link=mysql_connect("localhost", "root", "");
if(!$link){
die("ERROR".mysql_error());
}
mysql_select_db("databasename", $link);
$myusername=$_POST['myusername'];
$mypassword=$_POST['mypassword'];

$sql="SELECT * FROM table_name WHERE username='$myusername' and password='$mypassword'";
$result=mysql_query($sql);
$count=mysql_num_rows($result);

if($count==1){
header("location:login_success.php");
}
else {
echo "Wrong Username or Password";
}

basic codes try daw!

[merinoabs]

kni nga part bai,

$query=('SELECT * FROM USERS WHERE USERNAME = "$user" and PASSWORD = "$pass"')

to

$query=('SELECT * FROM USERS WHERE USERNAME = "'.$user.'" and PASSWORD = "'.$pass.'"')

[senpai91]

Warning: mysql_num_rows() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\test\retrieve.php on line 13
Wrong Username or Password

error....pero still naka encounter ko ani...bisag ok tanan statements nako...false gi hapong ang statement nako...

wa jud ko kasabot ug ngano...basic raman unta ni...

di man sad dako ang nag knowledge nako aning php...first time paman ni nako...
interesado jud ko makat-on ani...

[senpai91]

kni nga part bai,

$query=('SELECT * FROM USERS WHERE USERNAME = "$user" and PASSWORD = "$pass"')

to

$query=('SELECT * FROM USERS WHERE USERNAME = "'.$user.'" and PASSWORD = "'.$pass.'"')

dude tried it with the concatenate still the same..

[stealthghost]

^ dude I did what you post but...

+++++++++++++++++++++++++++++++

<?php
$con=mysql_connect("localhost", "root", "") or die(mysql_error());
$db=mysql_select_db("user", $con) or die(mysql_error());


$user=$_POST['user'];
$pass=$_POST['pass'];


$query=('SELECT * FROM USERS WHERE USERNAME = "$user" and PASSWORD = "$pass"') or die(mysql_error());
$result=mysql_query($query);


$count = mysql_num_rows($result);
if($count == 1)
{
echo"Success";
}
else
{
echo"FAILED TO LOGIN!";
}

?>

this is still not working...

ang db ani kay naa nay sud username ug password...pero still false gi hapon ang iyang statement...

ngano kaha ni?

based sa akong experience, the Command should be,

SELECT * FROM USERS WHERE USERNAME = 'username' and PASSWORD = 'password' ;
on a normal MySQL Query.

But for PHP Syntax,
$query=("SELECT * FROM USERS WHERE USERNAME = '$user' and PASSWORD = '$password' ")

Naa pa na'y "." something, naka limut ko sa right syntax, it's in my laptop and I'm too lazy to open it right now, dugay mag load.

Problem with count, since wala kaayu restriction siguro imu code, it's better/safe to put,

COUNT>=1

Logically, if COUNT==1, but if you have two results, then it will be false

[Klave]

$query=('SELECT * FROM USERS WHERE USERNAME = "$user" and PASSWORD = "$pass"')

When using single quoted string, there will be no parsing.
The string will be taken as it is.
So you are comparing your usernames and passwords to the values "$user" and "$pass" respectively.