If value is true show, if value is false show something else - php

Question

I have some file names stored in my table with a category_id for each image.
I have the gallery all on one page.
So my question is, if my category_id's value is 3 then I want it to show the image.
If the category_id is not 3 I want to display a message saying, "no content", or whatever.
Here is my code. Not sure where I went wrong. Hoping someone can give me a hand.
<?php
$newlogos = DB::getInstance()->query("SELECT `logo_id`, `creation`, `logo_name`, `category_id`, `org_img`, `new_img` FROM `logos` WHERE category_id = 3 ORDER BY `logo_id` DESC ");
$category_id = ('category_id' == 3);
if($category_id == false){ ?>
<div><h2>No Content</h2></div>
<?php }
if($category_id == true){
foreach($newlogos->results() as $nl){ ?>
//Do Stuff
I've tried a bunch of combinations like this for example:
$category_id = ('category_id' == 3);
if($category_id != 3){ ?>
<div><h2>No Content</h2></div>
<?php }
else{
//Do Stuff
Just can't seem to figure it out.

Answer 1

Your statement here is not correct:
<?php
$category_id = ('category_id' == 3);
?>
You are trying to compare a string with an integer and this always returns false.
Edit:
After you execute your query, you need to fetch the data from the database. After you have stored the data in a variable (e.g. $data), you can compare it like this:
<?php
$category_id = ((int)$data['category_id'] == 3);
?>
Edit: #cwd:
Here is one way to do it:
<?php
//$con is database connection object
$result = mysqli_query($con,"SELECT `logo_id`, `creation`, `logo_name`, `category_id`, `org_img`, `new_img` FROM `logos` WHERE category_id = 3 ORDER BY `logo_id` DESC LIMIT 0,1");
$row = mysqli_fetch_assoc($result);
$category_id = (int)$row['category_id'] == 3 ?true:false;
?>
This might help you.