MySQL/PHP Error!

[BadManSam]

Hi, when I try to use a SELECT statement in php, I get an error:

Resource id #18

<?php
include("include/session.php");
$get = mysql_query("SELECT value FROM settings WHERE id=2");
echo $get;
?>

The file session.php contains the database credentials .etc and it's correct.
But then when I try to echo $get, I get the error.

Why, I'm not sure and I'm still learning MySQL/PHP.

Please could someone help and thank you in advance for any replies.

[paulkd]

You just need to read a little more.

Here's one article from a google search. I've picked mysqli rather than mysql as mysql is so last year :-)

[n21115]

Hi,

When you store the results of an SQL query for example, you need to pass it onto either mysql_fetch_array or mysql_fetch_assoc, and, go through a loop such as foreach, for example to output / use the information retrieved.

Also, since it seems you're relatively new with working with arrays, here's a tip. When working with, or outputting arrays in PHP, you can't simply 'echo' it. You must use a function like print_r, or better yet, use a loop like foreach to navigate through the array.

• n21115

[Derice]

Using var_dump instead of echo if your intention is to check on the value of the variable. Change "echo $get;" to "var_dump($get);". Hope it helps!

For more information on var_dump check this out http://php.net/manual/en/function.var-dump.php