mysqli Object returned instead of result - php

Question

I am running
$this->db->query("SELECT `id` FROM $table WHERE $table.id ='$product_id'");
that should return 92 but its returning the below why?
object(CI_DB_mysqli_result)#150 (8) { ["conn_id"]=> object(mysqli)#16
(18) { ["affected_rows"]=> int(1) ["client_info"]=> string(6) "5.5.30"
["client_version"]=> int(50530) ["connect_errno"]=> int(0)
["connect_error"]=> NULL ["errno"]=> int(0) ["error"]=> string(0) ""
["field_count"]=> int(1) ["host_info"]=> string(25) "Localhost via
UNIX socket" ["info"]=> NULL ["insert_id"]=> int(0) ["server_info"]=>
string(10) "5.5.31-cll" ["server_version"]=> int(50531) ["stat"]=>
string(150) "Uptime: 106781 Threads: 14 Questions: 30097132 Slow
queries: 13 Opens: 1937675 Flush tables: 1 Open tables: 400 Queries
per second avg: 281.858" ["sqlstate"]=> string(5) "00000"
["protocol_version"]=> int(10) ["thread_id"]=> int(373292)
["warning_count"]=> int(0) } ["result_id"]=> object(mysqli_result)#161
(5) { ["current_field"]=> int(0) ["field_count"]=> int(1)
["lengths"]=> NULL ["num_rows"]=> int(1) ["type"]=> int(0) }
["result_array"]=> array(0) { } ["result_object"]=> array(0) { }
["custom_result_object"]=> array(0) { } ["current_row"]=> int(0)
["num_rows"]=> NULL ["row_data"]=> NULL}

Answer 1

It is returning mysqli_ object.So Try to get the result like
$query = $this->db->query("SELECT `id` FROM $table WHERE $table.id ='$product_id'");
$result = $query->result();
foreach($result as $row)
{
echo "Id is ".$row['id']."<br>";
}
And it is appreciable that you are using mysqli_* functions instead of deprecated mysql_* functions

Answer 2

It's returning a mysqli_result object, exactly as the manual says it does.
To get the actual id you need to call fetch_assoc() (or similar) on the object.
if ($result = $this->db->query("SELECT id FROM $table WHERE $table.id ='$product_id'")) {
/* fetch associative array */
while ($row = $result->fetch_assoc()) {
printf ("Fetched ID: %s\n", $row["id"]);
}
/* free result set */
$result->free();
}

Answer 3

Basically the statement:
$this->db->query("SELECT 'id' FROM $table WHERE $table.id ='$product_id'");
returns an object that can be used to extract the result set or table and assigned to a variable... so you need to create a variable and assign the result set to it as:
$mysqli = new mysqli("localhost","rinonymous","03318987165oo","rinonymous");
if ($mysqli->connect_errno) {
print_r($mysqli->connect_error);
exit();
}
$site_title = "Rinonymous";
$page_title = "";
$page_body = "";
#Page Setup
$query_page_info = "select * from pages where id = 1";
foreach ($mysqli->query($query_page_info) as $row) {
print_r($mysqli->query($query_page_info));
#query method returns an associate array
print_r($row);
$page_title = $row['title'];
$page_body = $row['body'];
}