Why do I keep getting this error message?

[Horhay5150]

I am admittedly a noob to PHP and mySQL, Have been running through some tutorials. Whenever I run the following script I get this error message: mysql_query() expects parameter 2 to be resource, boolean given.

Any help would be greatly appreciated.

<?php


$input = $_POST['Rating'];


//connect to database
$link = mysql_connect("localhost", "name", "pword"); 
$db = mysql_select_db("test", $link);


$query = "SELECT RatingID FROM rating WHERE rating.Rating ='$input'";
$result = mysql_query($query, $db) or die(mysql_error()); 

$row = mysql_fetch_array($result);
        echo "Rating ID - ". $row['RatingID'];
        


?>

[Atli]

Hey.

That means that your mysql_connect call is failing. When it is successful, it returns a resource that you can pass into the mysql_query function. However, when it is unsuccessful, it returns FALSE which, when passed into the mysql_query function, generates this error.

You should verify that the mysql_connect call was successful before trying to pass the $link to the mysql_query function.

For example:
[code=php]// You can do:
$link = mysql_connect(' host', 'usr', 'pwd');
if($link === false) {
die("MySQL connection could not be established: " . mysql_error());
}

// Or simply:
$link = mysql_connect(' host', 'usr', 'pwd') or die(mysql_error ());[/code]

[Horhay5150]

added the suggested code, error message is still the same. For some reason mysql_Query isnt happy with $db, as far as i have been able to tell though my syntax on both connect and db selection is correct, i'm stumped.

[Atli]

added the suggested code, error message is still the same.

That can't be possible. If you added either of those and the connection fails, the code should be stopped before it reached the query call.

Can you show us how you modified the code?

[Atli]

Ahhh ok. I see now. Never mind my previous post.

You are using the wrong variable in your mysql_query call. You should be using the result of the mysql_connect call, not the mysql_select_db call.

[code=php]// This
mysql_query($qu ery, $db)

// Should be
mysql_query($qu ery, $link)[/code]

[Horhay5150]

THANK YOU. That did the trick. Appreciate the help.