CODEWITHSUNDEEP
phpwordpressif-statementheader
JOKKER
JOKKER

Reputation: 542

Display header based on user role and page (Wordpress)

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:

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

Upvotes: 0

Views: 1650

Answers (2)

amarinediary
amarinediary

Reputation: 5449

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.

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: ''

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

Upvotes: 1

Paul Mitchell
Paul Mitchell

Reputation: 142

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
}

?>

Upvotes: 2

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