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

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  • lisles
    New Member
    • Jan 2010
    • 40

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

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

    Code:
    /* 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

    Code:
    $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

    Code:
    <?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>
    Last edited by Markus; Jan 15 '10, 09:34 AM. Reason: Added [code] tags
  • dgreenhouse
    Recognized Expert Contributor
    • May 2008
    • 250

    #2
    My first guess is that the server has a password set and you're not providing it in the code...

    Code:
    db_server = "localhost"; //Database server
    $db_user = "root"; //Database user name
    [B]$db_pass = ""; //Database server password - MISSING PASSWORD
    [/B]$db = "panchayatsgoa"; //Database Name

    Comment

    • lisles
      New Member
      • Jan 2010
      • 40

      #3
      any idea how i can do dat?

      Comment

      • Markus
        Recognized Expert Expert
        • Jun 2007
        • 6092

        #4
        Originally posted by lisles
        any idea how i can do dat?
        Huh? You do not know your own MySQL server password?

        Comment

        • nomad
          Recognized Expert Contributor
          • Mar 2007
          • 664

          #5
          to find your password look it up on privileges

          Comment

          • Nirmal Bage

            #6
            first Establish the Connection then only it will work

            Comment

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