mysql_query() expects parameter 2 to be resource, boolean given in [duplicate] - php

Question

This question already has answers here:
mysql_fetch_array()/mysql_fetch_assoc()/mysql_fetch_row()/mysql_num_rows etc... expects parameter 1 to be resource
(31 answers)
Closed 8 years ago.
I looked at the other answers to problems similar to mine but cant seem to solve this.
Here is the code.
$connection = mysql_connect("localhost","root","starwars");
$conn = mysql_select_db("project", $connection);
// This code assumes $itemID is set to that of
// the item that was just rated.
// Get all of the user's rating pairs
$sql = "SELECT DISTINCT r.itemID, r2.ratingValue - r.ratingValue
as rating_difference
FROM rating r, rating r2
WHERE r.userID=$userID AND
r2.itemID=$itemID AND
r2.userID=$userID;";
$db_result = mysql_query($sql, $conn);
echo "The result is {$db_result}";
$num_rows = mysql_num_rows($db_result)or die('Cannot Execute:'. mysql_error());
The error being displayed is:
Warning: mysql_query() expects parameter 2 to be resource, boolean
given in C:\xampp\htdocs\recomender\ratingfiles\class.rating.php on
line 177
Line 177 is
$db_result = mysql_query($sql, $conn);
And if I echo $conn it gives the value of "1" which I thought was equal to true, thus boolean, any ideas?

Answer 1

Pass $connection as the second parameter, not $conn.
You assign the result of mysql_select_db to $conn, and mysql_select_db returns true or false, not a connection resource.

Answer 2

This probably means $conn is false, meaning it didnt get setup correctly. You may want to check how you have set that up and ensure the database connection details are correct and the server this is running on can access the database server.
Take a look at the return values on this page:
http://php.net/manual/en/function.mysql-connect.php

Answer 3

change first 2 lines to
$conn = mysql_connect("localhost","root","starwars");
mysql_select_db("project", $conn);
or, better, just refrain from using connection variable at all
mysql_connect("localhost","root","starwars");
mysql_select_db("project");
...
$db_result = mysql_query($sql);

Answer 4

$db_result = mysql_query($sql, $connection);
Second argument to mysql_connect must have
The MySQL connection. If the link identifier is not specified, the last link opened by mysql_connect() is assumed. If no such link is found, it will try to create one as if mysql_connect() was called with no arguments. If no connection is found or established, an E_WARNING level error is generated.
http://php.net/manual/en/function.mysql-query.php