Strange output for simple php mysql search - php

Question

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;
?>

Answer 1

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.

Answer 2

Try something similar to this - after first "SELECT" query :
while($user_id_obj = mysql_fetch_object($user_id))
{
echo $user_id_obj->id;
}
The way you implemented leads to SQL Injection Attacks
SQL Injection Attacks Example

Answer 3

This could be possible in two ways.Which is usefull for you is depends on your requirements.
1.if your query contains a single value as a result then following code with changes in your code will be usefull for you.
<?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
$result_user = mysql_query("SELECT id FROM users WHERE email = '$query'") or die(mysql_error());
if (!$result_user) {
die('Could not query:' . mysql_error());
}
$user_id=mysql_result($result_user,0); // outputs first user's id
echo $user_id;
$result_accountuser = mysql_query("SELECT 'account_id' FROM accounts_users WHERE 'user_id' LIKE ('$user_id')") or die(mysql_error());
if (!$result_accountuser) {
die('Could not query:' . mysql_error());
}
$account_id=mysql_result($result_accountuser,0); // outputs first accounts_users's account_id
echo $account_id;
$result_account = mysql_query("SELECT 'account_name' FROM accounts WHERE 'id' LIKE ('$account_id')") or die(mysql_error());
if (!$result_account) {
die('Could not query:' . mysql_error());
}
echo mysql_result($result_account,0); // outputs first accounts's account_name
?>
2.Or your query contains more than one result or more than one rows than following changes in your code will help you
<?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
$result_user = mysql_query("SELECT id FROM users WHERE email = '$query'") or die(mysql_error());
while($row=mysql_fetch_array($result_user))
{
$user_id = $row['id'];
echo $user_id;
}
$result_accountuser = mysql_query("SELECT 'account_id' FROM accounts_users WHERE 'user_id' LIKE ('$user_id')") or die(mysql_error());
while($row=mysql_fetch_array($result_accountuser))
{
$account_id = $row['account_id'];
echo $account_id;
}
$result_account = mysql_query("SELECT 'account_name' FROM accounts WHERE 'id' LIKE ('$account_id')") or die(mysql_error());
while($row=mysql_fetch_array($result_account))
{
echo $row['account_name'];
}
?>