Not a valid argument?

D_tunisia

Can anyone please tell me why I get this message:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/.... on line 9

With this code:

<?php

$result = mysql_query("SELECT * FROM category_description, product_to_category, product_description WHERE category_description.name = $category[name] AND category_description.category_id = product_to_category.category_id AND product_to_category.product_id = product_description.product_id");

while($myrow = mysql_fetch_array($result)) {

$product_array[] = $product_array . ", " . $myrow['product_description.name'];	

}

?>

I have checked and double checked that the tables and attributes called are correct. Is there something I am missing with the query?

Thanks in advance for any help!!!

laserlight

Check your database connection and that the correct database was selected. If those are correct, then the error probably lies in your SQL statement.

jamesm6162

Start by setting
error_reporting = E_ALL
in your php.ini file. If there are any errors, then it will show them to you.

Then call

var_dump(mysql_error())

right after your call to mysql_query() to see if there were any errors.

I can't see much wrong with the query. Try cleaning up your code like e.g. placing the "name" inside single quotes:

$category['name']

Then in your WHERE clause, place single quotes in front and after all strings:

category_description.name =[B]'[/B]$category[name][B]'[/B]

bradgrafelman

jamesm6162 wrote:
like e.g. placing the "name" inside single quotes

Actually, doing so would cause a parse error; array indexes inside double-quoted strings should not be quoted, whereas they should be if they appear outside of a string (e.g. in concatenation).

jamesm6162 wrote:
Then in your WHERE clause, place single quotes in front and after all strings

James makes a good point - all strings must be quoted, just like in PHP.

Also, whenever you have a SQL problem, the usual approach to debugging a query is like so:

$query = 'SELECT foo FROM bar WHERE ...'; // store query in a variable
$exec = mysql_query($query) or die('MySQL error: ' . mysql_error() . '<br>Query: ' . $query);

That way, if the query fails, you'll get the error message from SQL as well as what your query looked like.

jamesm6162

bradgrafelman wrote:
Actually, doing so would cause a parse error; array indexes inside double-quoted strings should not be quoted

Yes Indeed, my mistake.

You have to actually place the entire variable inside {}-brackets.

That's the way I do it mostly