Display header based on user role and page (Wordpress) - php

Question

I have created multiple different headers as templates with Elementor.
I would like to display all of the different headers based on user role (Logged in/Logged out) and page.
I'm looking for a code snippet that I could easily customize to assign all of the different headers for different scenarios.
Could someone please create an example code snippet that would:
Display header A for Logged Out users on the entire
website, EXCEPT pages X and Y.
Display header B for Logged In users on the entire
website, EXCEPT pages X and Y.
Display header C for Logged Out users only on pages X and
Y.
Display header D for Logged In users only on pages X and
Y.
This way, people can easily copy the code and customize it to fit their needs.
EDIT
There's 2 places where I can create templates.
1st one is added by Elementor and is found in Admin > Templates > Saved Templates. Here I can create either section or page templates (Screenshot).
2nd one is added by my theme, OceanWP, and is found in Admin > Theme Panel > My Library. Here I can create only 1 type of template. The templates created here can later be assigned as custom headers or footers to individual pages or the entire website.
Are the templates created in these 2 places considered to be template parts? Is there a difference where I choose to create the header templates?
Here's a list of the header templates I have created:
Template title
Post ID
A
Main Header (Logged Out)
5448
B
Main Header (Logged In)
6714
C
Checkout Header (Logged Out)
6724
D
Checkout Header (Logged In)
3960
Here's the page I want to have a different header than the entire website:
Page title
Post ID
Slug
X
Checkout
18
checkout

Answer 1

Something like this should work:
<?php
if (! is_user_logged_in() && ! is_page(array( 'page-x-slug', 'page-y-slug' ))){
// display header A
}
if (is_user_logged_in() && ! is_page(array( 'page-x-slug', 'page-y-slug' ))){
// display header B
}
if (! is_user_logged_in() && is_page(array( 'page-x-slug', 'page-y-slug' ))){
// display header C
}
if (is_user_logged_in() && is_page(array( 'page-x-slug', 'page-y-slug' ))){
// display header D
}
?>

Answer 2

The following offer the same result as the previous answer but is minified and has less repetitiveness.
<?php
if ( is_page( [ 'page-x', 'page-y' ] ) )
if ( is_user_logged_in() )
get_header( 'B' ); //... header-B.php
else
get_header( 'A' ); //... header-A.php
else
if ( is_user_logged_in() )
get_header( 'D' ); //... header-D.php
else
get_header( 'C' ); //... header-C.php
?>
Following your comments
I'm guessing that what you refer as...
section templates
...are in fact templates part. Instead of using get_header( string $name ); you would then use get_template_part( string $slug, string $name = null );.
$slug and $name can be anything that you chose.
Source # https://developer.wordpress.org/reference/functions/get_template_part/
For example, section-A.php would be get_template_part( 'section', 'A' );.
<?php
//...
if ( is_user_logged_in() )
get_template_part( 'section', 'B' ); //... section-B.php
else
get_template_part( 'section', 'A' ); //... section-A.php
?>
In regards to specifying pages and templates. is_page() can take IDs, slugs or titles.
is_page( int|string|int[]|string[] $page = '' )
Parameter
Description
$page
(int|string|int[]|string[]) (Optional) Page ID, title, slug, or array of such to check against. Default value: ''
Source # https://developer.wordpress.org/reference/functions/is_page/
But you could also use other is_ function like is_search() or is_archives(), is_404()... etc.
Here is a complete list # https://codex.wordpress.org/Conditional_Tags
If you want to add multiple conditional statement you can just add elseif statements in-between.
<?php
if ( is_page( [ 'page-x', 'page-y' ] ) )
//...
elseif ( is_search() || is_archive() )
//...
else
//...
?>
If you want to get a better understanding of PHP operators, which are how conditional statements are built, take a look # https://www.w3schools.com/php/php_operators.asp