Warning: mysql_query() [function.mysql-query]: Access denied for user 'ODBC'@'localho

40 New Member

i have a page funtion.php which hs the function to connect to the db

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 /* Mysql Connection */

function connect(){

    global $db_server,$db_user,$db_pass,$db;//Global Values from the config.php

    $conn = @mysql_connect($db_server,$db_user,$db_pass) or die("Connection to Database Server Failed");

    @mysql_select_db($db) or die("Database Selection Failed");

    return $conn;

}
 
/* MYSQL QUERY FUNCTION */

function caseQuery($sqlString){

    $conn = connect();

    $rows = @mysql_query($sqlString,$conn);

    @mysql_close($conn);

    return $rows;

}
 
/* MYSQL INSERT QUERY FUNCTION */

function caseInsertQuery($sqlString){

    $conn = connect();

    $rows = @mysql_query($sqlString,$conn);

    $insertId = @mysql_insert_id($conn);

    @mysql_close($conn);

    return $insertId;

}

2)i have a page page config.php which has the following code

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 $db_server = "localhost"; //Database server

$db_user = "root";                //Database user name

$db_pass = ""; //Database server password

$db = "panchayatsgoa"; //Database Name
 
$filePath = $_SERVER['DOCUMENT_ROOT'].""; //Complete File Path on the webserver Without trailing slash
 
$downloadPath = "downloads"; //This should be in the root directory of the WEBSITE this is a Folder name
 
/* Default Session Time out */

$sessionTimeout = 3600; //Session time out should be specified in SECONDS;

3)i have a page login_attempts. php which hs thecode shown below and which gives me the warning
Warning: mysql_query() [function.mysql-query]: Access denied for user 'ODBC'@'localho st' (using password: NO) in C:\xampp\htdocs \panchayats\adm in\login_attemp t.php on line 24

Warning: mysql_query() [function.mysql-query]: A link to the server could not be established in C:\xampp\htdocs \panchayats\adm in\login_attemp t.php on line 24

There was an error in selecting login logs

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 <?php

require_once "../inc/functions.php";

sessionCheck();

?>

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">

<html xmlns="http://www.w3.org/1999/xhtml">

<head>

<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />

<title>Control Panel</title>

<link rel="stylesheet" type="text/css" href="admin.css" />

<script language="javascript" type="text/javascript" src="menu.js"></script>

</head>
 
<body>
 
<h3> Login Attempts</h3>

<?php

    $page = $_GET['page'];

    $records_per_page = 10;

    if (!ctype_digit($page)) $page=1;

    $offset = ($page-1) * $records_per_page;
 
    $getlogs = "SELECT * FROM login_attempts ORDER BY id desc LIMIT $offset, $records_per_page";

    $result = mysql_query($getlogs) or die ("<div style='padding:20px; font-weight:bold; color:#FF0000;'>There was an error in selecting login logs</div>");

    $rows = mysql_num_rows($result);

    if ($rows) {

        echo "<div style='padding-left:20px;'><table style='border-collapse:collapse; border-color:#000' border='1' cellspacing='0' cellpadding='5' width='400'>";

        echo "<tr><td width='100'>IP Address</td><td width='60'>Attempts</td><td width='150'>Time Stamp</td></tr>";

        for($i=1;$i<=$rows;$i++) {

            $res=mysql_fetch_array($result);

            echo "<tr><td>{$res['ipaddress']}</td><td>{$res['attempts']}</td><td>{$res['timestamp']}</td></tr>";

        }

        echo "</table></div>";

        //paging toolbar

        $count_result = mysql_query("SELECT COUNT(*) FROM login_attempts");

        $count_row = mysql_fetch_array($count_result);

        $count = $count_row["COUNT(*)"];        //fetch the total number of rows in the table

        $numofpages = ceil($count/$records_per_page);        // how many pages we have when using paging?

        echo "<div style='margin-left:20px;width:400px;'>";

        if ($numofpages > '1' ) { pagingScript("login_attempt.php", $page, $numofpages); }

        echo "</div>";

    } else {

        echo "<div style='margin-left:20px;'>No Login Attempts</div>";

    }

?>

</body>

</html>

Jan 15 '10 #1

Subscribe Reply

9136

dgreenhouse

250

Recognized Expert Contributor

My first guess is that the server has a password set and you're not providing it in the code...

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db_server = "localhost"; //Database server

$db_user = "root"; //Database user name
 $db_pass = ""; //Database server password - MISSING PASSWORD
  $db = "panchayatsgoa"; //Database Name

Jan 15 '10 #2

lisles

New Member

any idea how i can do dat?

Jan 15 '10 #3

Markus

6,050

Recognized Expert Expert

@lisles
Huh? You do not know your own MySQL server password?

Jan 15 '10 #4

nomad

664

Recognized Expert Contributor

to find your password look it up on privileges

Jan 18 '10 #5

Nirmal Bage

first Establish the Connection then only it will work

Sep 29 '10 #6

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