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Thread: PHP code included within html

  1. #1
    Join Date
    May 2011
    Posts
    7

    Default PHP code included within html

    I have a landing page in which there is an include of a php module. I know it is executing because it is delivering echo output to page. However, the mysql query that seems to work in the authentication module, when ported and replacing code in this included code generates the following nasty-grams.

    Warning: mysql_query() [function.mysql-query]: Access denied for user 'acunishi'@'localhost' (using password: NO) in /home1/acunishi/public_html/go/auth/birthdays_holidays-greetingsv2.php on line 29

    Warning: mysql_query() [function.mysql-query]: A link to the server could not be established in /home1/acunishi/public_html/go/auth/birthdays_holidays-greetingsv2.php on line 29
    Invalid query: Access denied for user 'acunishi'@'localhost' (using password: NO)

    Any thoughts?

    Here is the beginning of the code with name of database, table, username and password eliminated for security.

    <?php
    session_start();
    $host="localhost"; // Host name
    $username=""; // Mysql username
    $password=""; // Mysql password
    $db_name=""; // Database name
    $tbl_name=""; // Table name

    // Connect to server and select databse.
    $dblink = mysqli_connect($host,$username, $password,$db_name);
    //mysqli_select_db("$db_name")or die("cannot select DB");

    /* check connection */
    if (mysqli_connect_errno()) {
    printf("Connect failed: %s\n", mysqli_connect_error());
    exit();
    }
    // username and password sent from form
    $myusername=$_POST['myusername'];
    $mypassword=$_POST['mypassword'];

    // To protect MySQL injection (more detail about MySQL injection)
    $myusername = stripslashes($myusername);
    $mypassword = stripslashes($mypassword);
    $myusername = mysqli_real_escape_string($dblink,$myusername);
    $mypassword = mysqli_real_escape_string($dblink,$mypassword);
    //end of setup and database open

    $result3 = mysql_query("SELECT nickname, announce, lastlogin from '$db_name'.'$tbl_name' WHERE '$tbl_name'.'username' = '$myusername'");
    if (!$result3) {
    die('Invalid query: ' . mysql_error());
    }


    $num_rows = mysql_num_rows($result3);
    $row3 = mysql_fetch_assoc($result3);
    echo " \n <br>";
    echo $row3['nickname'.' '.'announce'.' '.'lastlogin'];
    echo " ";
    echo " \n";


    - Show quoted text -

  2. #2
    Join Date
    Nov 2006
    Location
    Sydney, Australia
    Posts
    4,951

    Default

    The most likely cause is that you left off your account name from the fromt of the database name and username. They should be written as

    accountname_databasename
    accountname_username

  3. #3
    Join Date
    May 2011
    Posts
    7

    Default account name and database name/username

    My account name is "acunishi". When I created the database called "loginsv4" I used "acunishi_loginv4" as the database name in the code.

    The database has a created user with admin priviledges that when appended to account name, was "acunishi_php" (an arbitrary user name selection). That username is in the code.

    So I don't understand your reply ... and they are the same database name as in accountname_databasename and accountname_username. That cam directly from the authentication module which is working.

    Clearly, I am not understanding your response. What am I missing? This worked the same way with a previous webhosting company (accountname_username and databasename), so I am familiar with how that works.

    Respectfully...

  4. #4
    Join Date
    Jan 2008
    Location
    Cheney, WA
    Posts
    16

    Default

    I noticed that you used mysqli to connect to the database and mysql for the query. Try using all mysqli or mysql functions without mixing them.

    Bill

  5. #5
    Join Date
    Feb 2006
    Location
    Florida, USA
    Posts
    1,505

    Default

    Hollenback is right, it's defaulting to "mysql_connect" with no arguments because it has nothing else to go on. Switch to whichever the majority of your code uses.
    If at first you don't succeed, try reading the instructions.
    semlar.com

  6. #6
    Join Date
    May 2011
    Posts
    7

    Default

    This is the code I use for the database and never have a problem for 5 years;
    function db()
    {
    $dbuser = "texleath_drmills";
    $dbserver = "localhost";
    $dbpass = "texcig";
    $dbname = "texleath_sales";

    //CONNECTION STRING
    $db_conn = mysql_connect($dbserver, $dbuser, $dbpass)
    or die ("UNABLE TO CONNECT TO DATABASE");
    mysql_select_db($dbname)
    or die ("UNABLE TO SELECT DATABASE");
    return $db_conn;
    }//end function db

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