Comment obtenir un identifiant de publication personnalisé à partir de slug?

Vouspouvez utiliser get_page_by_path() -ne vous laissezpastromperpar lenom,letroisième argumentest letype depublication:

 if ( $post = get_page_by_path( 'the_slug', OBJECT, 'post_type' ) )
    $id = $post->ID;
else
    $id = 0;
 

You can use get_page_by_path() - don't let the name fool you, third argument is the post type:

if ( $post = get_page_by_path( 'the_slug', OBJECT, 'post_type' ) )
    $id = $post->ID;
else
    $id = 0;

Si vous attendez quelquesjourset passez à Wordpress 4.4 qui sortira le8 décembre ( AFAIK ),vouspouvez utiliser lenouveauparamètre post_name__in dans WP_Query quiprend untableau de slugs

EXEMPLE

Si vous avezbesoin de l'objet depublication complet

 $args = [
    'post_type'      => 'my_custom_post_type',
    'posts_per_page' => 1,
    'post_name__in'  => ['post-slug']
];
$q = get_posts( $args );
var_dump( $q );
 

Si vousn'avezbesoin que de l'ID

 $args = [
    'post_type'      => 'my_custom_post_type',
    'posts_per_page' => 1,
    'post_name__in'  => ['post-slug'],
    'fields'         => 'ids' 
];
$q = get_posts( $args );
var_dump( $q );
 

If you wait a couple of days, and upgrade to Wordpress 4.4 which will be released the 8th of December (AFAIK), you can use the new post_name__in parameter in WP_Query which takes an array of slugs

EXAMPLE

If you need the complete post object

$args = [
    'post_type'      => 'my_custom_post_type',
    'posts_per_page' => 1,
    'post_name__in'  => ['post-slug']
];
$q = get_posts( $args );
var_dump( $q );

If you only need the ID

$args = [
    'post_type'      => 'my_custom_post_type',
    'posts_per_page' => 1,
    'post_name__in'  => ['post-slug'],
    'fields'         => 'ids' 
];
$q = get_posts( $args );
var_dump( $q );

Si vous voulezjuste l'identifiant de l'article,celafera l'affaireen une seule ligne.

 url_to_postid( site_url('the_slug') );
 

If you just want the post id this will do the trick in one line.

url_to_postid( site_url('the_slug') );