PHP-MySQL Database Connection

Question

I'm learning PHP with videos of lynda.com.I created a database called widget_corp in my localhost phpmyadmin panel and I wrote these code block

    <?php
/* 1.Create a database connection */
$connection = mysql_connect("localhost", "root", "*");
if(!$connection){
die("Database connection failed: " .mysql_error());
}

/* 2. Select a database to use */
$db_select = mysql_select_db("widget_corp", $connection);
if(!$db_select)
{
die("Database selection failed: " . mysql_error());
}
?>

    <html>
    <head>
    <title> Connection To the Database </title>
    </head>
    <body>
    <?php

//3. Perform database query
$result = mysql_query("SELECT * FROM subjects",$connection);
if(!$result)
{
die("Database query failed: " .mysql_error());
}

//4. Use returned data

while($row = mysql_fetch_array($result));
{
echo $row["menu_name"]." ".$row["position"]."<br/>";
}
?>
</body>
</html>
<?php
//5. Close connection
mysql_close($connection);
?>

I always get this error type:

Object not found!

The requested URL was not found on this server. If you entered the URL manually please check your spelling and try again.

If you think this is a server error, please contact the webmaster.

Error 404

localhost
Apache/2.4.4 (Unix) PHP/5.5.3 OpenSSL/1.0.1e mod_perl/2.0.8-dev Perl/v5.16.3

How can I overcome this issue? thanks all

Its got to be with the location of your file. The code looks ok. try place the code directly in the www root and browse directly to localhost/conn.php — Rob
– Rob, Commented Nov 24, 2013 at 9:11
For what it is worth, I agree with Rob. Try giving whoever hosts your service a call. It may be configuration as you are getting an apache error. Do you get anything if you run <?PHP phpinfo(); ?> if not then conifguration of PHP is the problem. — Robert Seddon-Smith
– Robert Seddon-Smith, Commented Nov 24, 2013 at 9:41

ashatte · Accepted Answer · 2013-11-26 08:33:18Z

5

You made a very small mistake. You used a semicolon in while statement, so just remove it. Then it will work fine.

Use returned data:

while($row = mysql_fetch_array($result))

Not:

 while($row = mysql_fetch_array($result));

edited Nov 26, 2013 at 8:33

ashatte

5,5388 gold badges41 silver badges51 bronze badges

answered Nov 24, 2013 at 11:19

harry

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