Can't get output in php how to fix it?

Question

PHP

Guys can anyone help
I'm trying to make a website and there are a number of problems.
For example, I want to display pictures on the site from the database, I do everything like this

$sql_img = $db->super_query("SELECT * FROM imgtop");
foreach($sql_img as $q)
  $img .= '<li  style="background-image: url({theme}/img/'.$sql_img['name'].');" class="overlay"></li> ';	
$tpl->set('{img}', $img);

There are 4 pictures in the database in the name table, the names of the pictures are indicated. It renders the image like this:

<li  style="background-image: url(/system_sain/lk_system/system_tpl/komp/img/slider_1.jpg);" class="overlay"></li> 
<li  style="background-image: url(/system_sain/lk_system/system_tpl/komp/img/slider_1.jpg);" class="overlay"></li>

For some reason, he himself immediately outputs 2 pictures one and the same.
How can I fix this so that he would display one picture and not 2 at once one and the same?
How can I make it output from mysql all the pictures that are there and they were shown?
I already created threads with this question. Please read the answers here How to fix in php?
In order not to ask the same questions and write such answers.

Answer 1

1) you can write what var_dump($sql_img); will display;
2) it is more logical to think that it would be more correct to write instead of $sql_img['name'] -> $q['name']

Answer 2

$sql_img = $db->super_query("SELECT * FROM imgtop limit 2", true);
print_r($sql_img);
exit;

what does it show?
Outputs an array like this
Array ( [0] => Array ( [id] => 1 [name] => slider_1.jpg ) [1] => Array ( [id] => 2 [name] => slider_2.jpg ) )
sandbox.onlinephpfunctions.com/code/9f43dc6039aa53...
result:
someone piz ... tryndit, and most likely not an interpreter)