[SOLVED] Login not working :(

[pure_skill_2000]

Hi

 

My login form wont work for some reason, any ideas?

 

<form name="authenticate" method="post" action="wk9ex2.php">
<input name="login id" type="text" value="loginid" size="20"/><br >
<input name="password" type="text" value="password" size="20"/><br>
<input type="submit" name="submit" value="submit"/>
<input type="reset" name="reset" value="reset"/>

<?php

//Database Information

$dbhost = "";
$dbname = "";
$dbuser = "";
$dbpass = "";

//Connect to database

mysql_connect ( $dbhost, $dbuser, $dbpass)or die("Could not connect: ".mysql_error());
mysql_select_db($dbname) or die(mysql_error());

$sql="SELECT loginid FROM week9 WHERE loginid='".$loginid.”’ and password=’”.$password.”’”;
$r = mysql_query($sql);
if(!$r) {
$err=mysql_error();
print $err;
exit();
}
if(mysql_affected_rows()==0){
print "no such login in the system. please try again.";
exit();
}
else{
echo loggedin.php
//proceed to perform website’s functionality – e.g. present information to the user
}
?>

[shocker-z]

HOPE THATS NOT YOUR ACTUAL USER AND PASS!!!!?

 

you haven't defined $loginid and $password

 

try this

 

<form name="authenticate" method="post" action="wk9ex2.php">
<input name="login id" type="text" value="loginid" size="20"/><br >
<input name="password" type="text" value="password" size="20"/><br>
<input type="submit" name="submit" value="submit"/>
<input type="reset" name="reset" value="reset"/>

<?php

//Database Information

$dbhost = "";
$dbname = "";
$dbuser = "";
$dbpass = "";

//Connect to database

mysql_connect ( $dbhost, $dbuser, $dbpass)or die("Could not connect: ".mysql_error());
mysql_select_db($dbname) or die(mysql_error());

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

$sql="SELECT loginid FROM week9 WHERE loginid='".$loginid.”’ and password=’”.$password.”’”;
$r = mysql_query($sql);
if(!$r) {
$err=mysql_error();
print $err;
exit();
}
if(mysql_affected_rows()==0){
print "no such login in the system. please try again.";
exit();
}
else{
echo loggedin.php
//proceed to perform website’s functionality – e.g. present information to the user
}
?>

 

 

Regards

Liam

[pure_skill_2000]

This produces an

 

syntax error, unexpected '='

 

Error on line 33

 

[shocker-z]

What is line 33 in your code as i make it out to be

 

exit();

 

but do you have more code at the top or somthing as i cant see an issue with the code i inserted.

 

 

Liam

[pure_skill_2000]

$sql="SELECT loginid FROM week9 WHERE loginid='".$loginid.”’ and password=’”.$password.”’”;

[corbin]

Try $sql="SELECT loginid FROM week9 WHERE loginid='{$loginid}' and password='{$password}'";

[shocker-z]

use this

 

$sql="SELECT loginid FROM week9 WHERE loginid='".$loginid."' and password='".$password."'";

 

You used the wrong " and ' simbols..

 

Could possibly have been just when you pasted or editor you used..

 

Regards

Liam

[pure_skill_2000]

Below is my code atm, it still inst working though, when I click to send the information it just refreshes the page instead of going to the logged in page

 

 

Any ideas?

 

<form name="authenticate" method="post" action="wk9ex2.php">
<input name="login id" type="text" value="loginid" size="20"/><br >
<input name="password" type="text" value="password" size="20"/><br>
<input type="submit" name="submit" value="submit"/>
<input type="reset" name="reset" value="reset"/>

<?php

//Database Information


//Connect to database

mysql_connect ( $dbhost, $dbuser, $dbpass)or die("Could not connect: ".mysql_error());
mysql_select_db($dbname) or die(mysql_error());

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

$sql="SELECT username FROM week9 WHERE username='".$loginid."' and password='".$password."'";
$r = mysql_query($sql);
if(!$r) {
$err=mysql_error();
print $err;
exit();
}
if(mysql_affected_rows()==0){
print "no such login in the system. please try again.";
exit();
}
else{
echo loggedin.php;
//proceed to perform website’s functionality – e.g. present information to the user
}
?>

[chris_rulez001]

Below is my code atm, it still inst working though, when I click to send the information it just refreshes the page instead of going to the logged in page

 

 

Any ideas?

 

<form name="authenticate" method="post" action="wk9ex2.php">
<input name="login id" type="text" value="loginid" size="20"/><br >
<input name="password" type="text" value="password" size="20"/><br>
<input type="submit" name="submit" value="submit"/>
<input type="reset" name="reset" value="reset"/>

<?php

//Database Information


//Connect to database

mysql_connect ( $dbhost, $dbuser, $dbpass)or die("Could not connect: ".mysql_error());
mysql_select_db($dbname) or die(mysql_error());

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

$sql="SELECT username FROM week9 WHERE username='".$loginid."' and password='".$password."'";
$r = mysql_query($sql);
if(!$r) {
$err=mysql_error();
print $err;
exit();
}
if(mysql_affected_rows()==0){
print "no such login in the system. please try again.";
exit();
}
else{
echo loggedin.php;
//proceed to perform website’s functionality – e.g. present information to the user
}
?>

 

this section of your code might be causing a problem:

 

echo loggedin.php;

 

try:

 

include("loggedin.php");

 

does your code work after you made the alteration? to the bit i pointed out

[pure_skill_2000]

Hi I made that change but it didnt seem to make any difference

 

The form just reposts a blank empty form, it does not seem to check the login and proceed to the logged in page

[juapo2]

Check all your code from all scripts, not getting MySQL error?

 

Yes, try

<?php include("loggedin.php");

 

use the

<?php 

statement

 

instead of

<?

 

tell me if you get any result

[shocker-z]

use this

 

<?php

//Database Information


//Connect to database

mysql_connect ( $dbhost, $dbuser, $dbpass)or die("Could not connect: ".mysql_error());
mysql_select_db($dbname) or die(mysql_error());

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

$sql="SELECT username FROM week9 WHERE username='".$loginid."' and password='".$password."'";
$r = mysql_query($sql);
if(!$r) {
  $err=mysql_error();
  print $err;
  exit();
}
if(mysql_num_rows($r)==0){
  print "no such login in the system. please try again.";
?>
<form name="authenticate" method="post" action="wk9ex2.php">
<input name="login id" type="text" value="loginid" size="20"/><br >
<input name="password" type="text" value="password" size="20"/><br>
<input type="submit" name="submit" value="submit"/>
<input type="reset" name="reset" value="reset"/>
<?php
} else {
  include('loggedin.php');
//proceed to perform website&#8217;s functionality &#8211; e.g. present information to the user
}
?>

 

I have made it so your form will show when not validated and when validated you will include the loggedin.php file. I changed mysql_affected_rows() to mysql_num_rows($row) as mysql_affected_rows is only used after updating a table, this will always be 0 when selecting from a table.

 

Regards

Liam