Cannot Connect to Database

[tymierb]

I am writing an application to do annual reviews. I cannot get my script to work. Whenever it runs I get the following error:

$username='far'; $password='c00lfarus3r@'; $database='far';
Warning: mysql_connect() [function.mysql-connect]: Access denied for user 'www-data'@'localhost' (using password: NO) in /var/www/FARv2/scripts/insertsection1.php on line 18
Unable to select database.

 

This does not make sense to me. It has the username and password variables and does not seem to be using them. Who is user www-data? I am pretty new to php so I don't know where to start.

 

Here are all of the files:

 

Input Page - section1.php:

<?php include('./includes/doc_top.php');?>

    <div class="container">
        <form action="scripts/insertsection1.php" method="POST">
            <h1>Teaching/Advising</h1>
            
            <div class="entry_box">
                <h4>1.A. Courses taught including the summer term.</h4>
                <textarea name="_1a"></textarea><br>
            </div>
            <strong>Note:</strong> The information provided below is automatically generated:
            
            <table>
                <tr>
                    <td>Course</td><td>Title</td><td>Term</td>
                </tr>
            </table>
            
            <div class="entry_box">
                <h4>1.B. Independent Study Supervision</h4>
                <textarea name="_1b"></textarea>
            </div>
            
            <h3>1.C. Other Academic Assignments</h3>
            
            <div class="entry_box">
                <h4>1. Undergraduate Advising</h4>
                <textarea name="_1c1"></textarea>
            </div>
            
            <div class="entry_box">
                <h4>2. Senior Design Committee</h4>
                <textarea name="_1c2"></textarea>
            </div>
            
            
            <div class="entry_box">
                <h4>1.D. Teaching Awards and Recognition</h4>
                <textarea name="_1d"></textarea>
            </div>
            
            
            <div class="entry_box">
                <h4>1.E. Course-Faculty Survey Results</h4>
                <textarea name="_1e"></textarea><br>
            </div>
            <strong>Note:</strong> The information provided below is automatically generated:
            
            <table>
                <tr>
                    <td>Course</td><td>Term</td><td>Course Rating</td><td>Faculty Rating</td><td># of Student Responses</td>
                </tr>
            </table>
            
            <div class="entry_box">
                <h4>1.F. Course/Lab Development and Major Revision Activity</h4>
                <textarea name="_1f"></textarea><br>
            </div>
            
            <h3>1.G. Student Design Projects</h3>
            <div class="entry_box">
                <h4>1.G.1. Freshman Design Supervision</h4>
                <textarea name="_1g1"></textarea>
            </div>
            
            <div class="entry_box">
                <h4>1.G.2. Senior Design Supervision</h4>
                <textarea name="_1g2"></textarea>
            </div>
            
            <div class="entry_box">
                <h4>1.H. Examples of Innovation in Teaching</h4>
                <textarea name="_1h"></textarea>
            </div>
            
            <div class="entry_box">
                <h4>1.I. E-Learning and Distance Learning</h4>
                <textarea name="_1i"></textarea>
            </div>
            
            <div class="entry_box">
                <h4>1.J. Other Teaching Related Activities (UNIV 101, Honors Seminar)</h4>
                <textarea name="_1j"></textarea>
            </div>
            <input type="submit" value="Submit Secion 1">
        </form>
    </div>
    </body>
</html>

 

Input Script - insertsection1.php

 

<?php

include('./../includes/db_connect.php');

$_1a=$_POST['_1a'];
$_1b=$_POST['_1b'];
$_1c1=$_POST['_1c1'];
$_1c2=$_POST['_1c2'];
$_1d=$_POST['_1d'];
$_1e=$_POST['_1e'];
$_1f=$_POST['_1f'];
$_1g1=$_POST['_1g1'];
$_1g2=$_POST['_1g2'];
$_1h=$_POST['_1h'];
$_1i=$_POST['_1i'];
$_1j=$_POST['_1j'];

mysql_connect(localhost,$database,$password);
@mysql_select_db(far) or die ("Unable to select database.");

$query = "INSERT INTO far VALUES ('$_1a','$_1b','$_1c1','$_1c2','$_1d','$_1e','$_1f','$_1g1','$_1g2','$_1h','$_1i','$_1j')";
mysql_query($query);

mysql_close();

?>

 

Output Page - output.php

<?php

include('./../includes/db_connect.php');

mysql_connect(localhost,$username,$password);
@mysql_select_db($database) or die("Unable to select database");
$query="SELECT * FROM section1";
$result=mysql_query($query);

$num=mysql_numrows($result);

mysql_close();

echo "<b><center>Database Output</center></b>";

$i=0;
while($i < $num) {
    $_1a=mysql_result($result,$i,"_1a");
    $_1b=mysql_result($result,$i,"_1b");
    $_1c1=mysql_result($result,$i,"_1c1");
    $_1c2=mysql_result($result,$i,"_1c2");
    $_1d=mysql_result($result,$i,"_1d");
    $_1e=mysql_result($result,$i,"_1e");
    $_1f=mysql_result($result,$i,"_1f");
    $_1g1=mysql_result($result,$i,"_1g1");
    $_1g2=mysql_result($result,$i,"_1g2");
    $_1g=mysql_result($result,$i,"_1h");
    $_1i=mysql_result($result,$i,"_1i");
    $_1j=mysql_result($result,$i,"_1j");
}
?>


<h2>1.A. Courses taught including the Summer Term</h2>
<div class="output">
    <?php echo $_1a;?>
</div>

<h2>1.B. Independent Study Supervision
<div class="output">
    <?php echo $_1b;?>
</div>

<h2>1.C. Other Academic Assignments</h2>
<h3>1. Undergraduate Advising</h3>
<div class="output">
    <?php echo $_1c1;?>
</div>
<h3>2. Senior Design Committee</h3>
<div class="output">
    <?php echo $_1c2;?>
</div>

<h2>1.D Teaching Awards and Recognition</h2>
<div class="output">
    <?php echo $_1d;?>
</div>

<h2>1.E. Course-Faculty Survey Results
<div class="output">
    <?php echo $_1e;?>
</div>

<h2>1.F. Course/Lab Development and Major Revision Activity</h2>
<div class="output">
    <?php echo $_1f;?>
</div>

<h2>1.G. Student Design Projects</h2>
<h3>1. Freshman Design Supervision</h3>
<div class="output">
    <?php echo $_1g1;?>
</div>

<h3>2. Senior Design Supervision</h3>
<div class="output">
    <?php echo $_1g2;?>
</div>

<h2>1.H. Examples of Innovation in teaching
<div class="output">
    <?php echo $_1h;?>
</div>

<h2>1.I. E-Learning and Distance Learning
<div class="output">
    <?php echo $_1i;?>
</div>

<h2>1.J. Other Teaching Related Activities (UNIV 101, Honors Seminar)</h2>
<div class="output">
    <?php echo $_1j;?>
</div>

 

Usernames and Passwords - db_connect.php:

$username='far';
$password='c00lfarus3r@';
$database='far';

[Kaboom]

Is this on a webserver or your localhost? Like if you buy hosting from a site like hostgator, they give you a username like kaboom then on your connect tables you need to put kaboom_far and the same username and database is a problem ...

[dgoosens]

try

mysql_connect('localhost',$username,$password);

 

with the quotes

[Kaboom]

Or this may work

 

mysql_connect('localhost',$database,$username,$password);

[tymierb]

No go. I added the quotes around localhost and get the same error.

 

I added $database to the string like Kaboom suggested and it gave this error which i guess is pretty much the same:

$username='far'; $password='c00lfarus3r'; $database='far';
Warning: mysql_connect() [function.mysql-connect]: Access denied for user 'www-data'@'localhost' (using password: NO) in /var/www/FARv2/scripts/insertsection1.php on line 18
Unable to select database.

 

@Kaboom > this is my own localhost so I control all of the accounts that can access it. Any more suggestions?

[dgoosens]

Warning: mysql_connect() [function.mysql-connect]: Access denied for user 'www-data'@'localhost' (using password: NO) in /var/www/FARv2/scripts/insertsection1.php on line 18

Unable to select database.

makes me think you have not included your db_connect.php file correctly... Check this...

Maybe try by coding the $username, $password etc. directly in the file...

[PFMaBiSmAd]

The posted error is occurring in insertsection1.php, in the following line of code -

 

mysql_connect(localhost,$database,$password);

 

The second parameter is NOT the database name, it is the user name. I'm going to guess that the 'user' portion of the error message is actually your database name. When an error occurs, read it, it contains important clues and also read the actual code that is producing the error.