CODEWITHSUNDEEP
phpmysqldatabase-connection
user3026894
user3026894

Reputation: 503

PHP-MySQL Database Connection

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

Upvotes: 1

Views: 2120

Answers (1)

harry
harry

Reputation: 1007

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));

Upvotes: 5

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