如果父页面具有特定模板，则显示ACF

我正在尝试创建一个新的ACF规则，以便在父页面具有特定模板名称时显示字段。以下是我当前的尝试：

add_filter(\'acf/location/rule_types\', \'acf_location_rules_types\');
function acf_location_rules_types( $choices ) {

    $choices[\'Parent\'][\'parent_template\'] = \'Parent Template\';

    return $choices;

}

add_filter(\'acf/location/rule_values/parent_template\', \'acf_location_rules_values_parent_template\');
function acf_location_rules_values_parent_template( $choices ) {

    $templates = get_page_templates();

    if ( $templates ) {
        foreach ( $templates as $template_name => $template_filename ) {

            $choices[ $template_name ] = $template_name;

        }
    }

    return $choices;
}

add_filter(\'acf/location/rule_match/parent_template\', \'acf_location_rules_match_parent_template\', 10, 3);
function acf_location_rules_match_parent_template( $match, $rule, $options ) {

    $selected_template = $rule[\'value\'];

    global $post;
    $template = get_page_template_slug( $post->post_parent );

    if( $rule[\'operator\'] == "==" ) {

        $match = ( $selected_template == $template );

    } elseif($rule[\'operator\'] == "!=") {

        $match = ( $selected_template != $template );

    }

    return $match;
}

我认为问题在于我试图为当前页面获取父页面模板的方式。我甚至可以在函数内部的挂钩函数中获取父页面模板。php？

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