Here is the error message:
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in D:\public_html\mysql_test.php on line 15

My local server setup:

• Apache 2.2.9.
• PHP 5.2.6.
• MySQL 5

Here is the tutorial I have been following.
Everything has gone smoothly up until now.
http://www.bicubica.com/apache-php-mysql/
I am stuck on Configuring PHP to work with MySQL.

Here is the PHP used:

PHP Code:

<?php

# Define MySQL Settings
define("MYSQL_HOST", "localhost");
define("MYSQL_USER", "root");
define("MYSQL_PASS", "password");
define("MYSQL_DB", "test");

$conn = mysql_connect("".MYSQL_HOST."", "".MYSQL_USER."", "".MYSQL_PASS."") or die(mysql_error());
mysql_select_db("".MYSQL_DB."",$conn) or die(mysql_error());

$sql = "SELECT * FROM test";
$res = mysql_query($sql);

while ($field = mysql_fetch_array($res))
{
$id = $field['id'];
$name = $field['name'];

echo 'ID: ' . $field['id'] . '<br />';
echo 'Name: ' . $field['name'] . '<br /><br />';
}

?>

Thank!

assuming it connects to the database correctly, that error always occurs for me when I have a table typed wrong. Make sure the table 'test' actually exists.

Checked and double checked on the db name and table name. The error is referring to line 15.

Line 15: while ($field = mysql_fetch_array($res))

change the one line to...

PHP Code:

mysql_query($sql)or die(mysql_error()); 


and see what it outputs

Interesting. I get the following message, Table 'test.test' doesn't exist
Why is it calling it "test. test"? The database name is "test".

'test.test' means 'table with name 'test' located in database with name 'test''

Using this syntax you can select data from tables located in different databases by single query (using joins), assuming that you have enough privileges to access those other databases.

Your error message shows quite clearly that a table that you try to select from does not exist. In other words, there is no table 'test' in the 'test' database.

Found the problem. Was selecting the wrong table name.

This...
$sql = "SELECT * FROM test";

should have been...
$sql = "SELECT * FROM name";

Thanks for all your help!