block_template_part( string $part )

Print a template-part.

Contents

Parameters

$part

(string) (Required) The template-part to print. Use "header" or "footer".

Top ↑

Source

File: wp-includes/block-template-utils.php 

function block_template_part( $part ) {
	$template_part = get_block_template( get_stylesheet() . '//' . $part, 'wp_template_part' );
	if ( ! $template_part || empty( $template_part->content ) ) {
		return;
	}
	echo do_blocks( $template_part->content );
}

Expand full source code Collapse full source code View on Trac View on GitHub

Top ↑

Top ↑

Uses

Uses
Uses Description
wp-includes/block-template-utils.php: get_block_template()

Retrieves a single unified template object using its id.
wp-includes/blocks.php: do_blocks()

Parses dynamic blocks out of post_content and re-renders them.
wp-includes/theme.php: get_stylesheet()

Retrieves name of the current stylesheet.

Top ↑

Changelog

Changelog
Version Description
5.9.0 Introduced.

Top ↑

User Contributed Notes

  1. Skip to note 1 content
    You must log in to vote on the helpfulness of this noteVote results for this note: 1You must log in to vote on the helpfulness of this note
    Contributed by Carolina Nymark —

    Note that this outputs the content of the template part, and does not use any parameters added to the template part itself.

    In other words,
    If you have a template part, and you are adding it inside a HTML block template in a block theme with these parameters:

    wp:template-part {"slug":"footer","tagName":"footer","className":"site-footer"}

    Using block_template_part() to output the same template part in a PHP template, does not output the wrapping footer element and the site-footer CSS class.

You must log in before being able to contribute a note or feedback.