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Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given on line 56 Notice: Undefined va

blackpendant

By blackpendant
in PHP Coding Help

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blackpendant Newbie

blackpendant

Posted
Posted

sir can can u fix mine too.? I'm new to this language  : index.php

 

'<link rel="stylesheet" type="text/css" href="css/style.css" />
<?php include('dbcon.php');
include('header.php');
 ?>
</head>
<body>
<div class="navbar navbar-fixed-top">
<div class="navbar-inner">
<div class="container">
    
<a class="brand">
<img src="images/chmsc.png" width="60" height="60">
  </a>
<a class="brand">
<h2>SSG ELECTRONIC VOTING SYSTEM</h2>
<div class="chmsc_nav"><font size="4" color="white">Sibonga Community College</font></div>
  </a>


<?php include('head.php'); ?>


</div>
</div>
</div>
<div class="wrapper_admin">
</br>
</br>
</br>
<div id="element" class="hero-body-index">


<p><font color="white"><h2>Admin Login</h2></font></p>


<form method="POST" >
<table>
    <tr><td><font color="white">UserName:</font>  </td><td><input type="text"  name="UserName" class="UserName_hover"></td></tr>
<tr><td>...<td></tr>
    <tr><td><font color="white">Password:</font>  </td><td><input type="Password" name="Password" class="Password_hover"></td></tr>
<tr><td>...<td></tr>
<tr><td></td><td>
<button class="btn btn-primary" name="Login"><i class="icon-ok icon-large"></i> Login</button>
</td></tr>
<tr><td>
</td><tr>
</form>
</table>


</br>
<div class="error">
<?php


if (isset($_POST['Login'])){


$UserName=$_POST['UserName'];
$Password=$_POST['Password'];


$login_query=mysql_query("select * from users where UserName='$UserName' and Password='$Password'");
$row=mysql_fetch_array($login_query);
$f=$row['FirstName'];
$l=$row['LastName'];


if ($count > 0){
session_start();
$_SESSION['id']=$row['User_id'];
$_SESSION['User_Type']=$row['User_Type'];
$type=$row['User_Type'];


mysql_query("INSERT INTO history (data,action,date,user)VALUES('$f $l', 'Login', NOW(),'$type')")or die(mysql_error());




header('location:home.php');
}else{
?>
    <div class="alert alert-error">
    <button class="close" data-dismiss="alert">×</button>
   Please check your UserName and Password
    </div>
<?php } 


}


?> 
</div>
</div>
</br>
</br>
</br>
</br>
</br>


<?php include('footer3.php')?> 
</div>


    </body>


</html>
Muddy_Funster Regular Member

Muddy_Funster

Posted
Posted

please use code tags around your code when posting it here, it makes it much easier to read.

 

Your error is coming from the fact that your query is failing. change your query lines to the following to better see what the problem is:

$sql = "select * from users where UserName='$UserName' and Password='$Password'";
$login_query=mysql_query($sql) or die("Query Failed!<br><br>The server Reported the following : <br>".mysql_error()."<br><br> when trying to run : <br>$sql");
$row=mysql_fetch_array($login_query);
blackpendant Newbie

blackpendant

Posted
  • Author
Posted

 

please use code tags around your code when posting it here, it makes it much easier to read.

 

Your error is coming from the fact that your query is failing. change your query lines to the following to better see what the problem is:

$sql = "select * from users where UserName='$UserName' and Password='$Password'";
$login_query=mysql_query($sql) or die("Query Failed!<br><br>The server Reported the following : <br>".mysql_error()."<br><br> when trying to run : <br>$sql");
$row=mysql_fetch_array($login_query);

thank you for replying sir, i'll give it a try.. uhmm it works but what will i do sir please help! :confused:  it says..

Query Failed!

 

The server Reported the following : 

No database selected

 

when trying to run : 

select * from users where UserName='admin' and Password='admin

 

Can you Fix this for me sir please.!! :(

Yohanne Regular Member

Yohanne

Posted
Posted

Make sure first, that you are connected to your database. 

$sql("SELECT * FROM users WHERE UserName = '$UserName' && Password = '$Password'");
$login_query = mysql_query($sql) or die("Query Failed :$sql ".mysql_error());
	
				
$rows = mysql_num_rows($login_query);
if($rows == 1)
{
echo 'you got login';
}
else
{
echo 'you got wrong user and password';
}
.josh Advanced Member

.josh

Posted
Posted

It looks like you are probably connecting to mysql in dbcon.php but you aren't actually selecting the database.  Make sure you have a line in your code or include that looks something like this: mysql_select_db('your db name here');

blackpendant Newbie

blackpendant

Posted
  • Author
Posted

It looks like you are probably connecting to mysql in dbcon.php but you aren't actually selecting the database.  Make sure you have a line in your code or include that looks something like this: mysql_select_db('your db name here');

this is my dbcon.php <?php mysql_select_db('anon',mysql_connect('localhost','root','')); ?>

Muddy_Funster Regular Member

Muddy_Funster

Posted
Posted

Split that line up, and add some "or die"s to it so you can find where the problem is:

<?php
$con = mysql_connect('localhost', 'root', '') or die("Can't connect to database");
$db = mysql_select_db('annon', $con) or die("database not found on server, or insufficient permissions set for current account");
?>

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