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Home/ Questions/Q 7875431
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Editorial Team
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Editorial Team
Asked: June 3, 20262026-06-03T02:58:50+00:00 2026-06-03T02:58:50+00:00

I am trying to Build a simple search that first grabbs ‘query’ from a

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I am trying to Build a simple search that first grabbs ‘query’ from a form passed from a HTML form through the url to this script. Once I run the script I get the output: Resource id #140Resource id #141Resource id #142. Why am I getting this output and what does it mean?

Side note I am just using the “echo” as a way to see the output of each variable.

<?php 
//connect to database
mysql_connect("localhost", "user", "password") or die("Error connecting to database: " .mysql_error());

mysql_select_db("dataBase") or die(mysql_error());

?>
<?php
$query = $_GET['query']; 
// gets value sent over search form

$user_id = mysql_query("SELECT id FROM users WHERE email = '$query'") or die(mysql_error());
echo $user_id;

$account_id = mysql_query("SELECT 'account_id' FROM accounts_users WHERE 'user_id' LIKE ('$user_id')") or die(mysql_error());
echo $account_id;

$user_name = mysql_query("SELECT 'account_name' FROM accounts WHERE 'id' LIKE ('$account_id')") or die(mysql_error());
echo $user_name;    
?>
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  1. Editorial Team
    Editorial Team
    2026-06-03T02:58:51+00:00Added an answer on June 3, 2026 at 2:58 am

    This is not the way to print the results. The method mysql_query returns a resource that you have to use within a loop to actually print the results. For instance, loop at the second example in the official doc page.

    P.S. $query = $_GET['query']; using this statement you could have Sql injections problems.

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