如何将链接项目添加到WooCommerce我的帐户菜单中的另一页

Question

我正在尝试将链接项目添加到WooCommerce My帐户菜单中的另一个页面(我的BuddyPress页面(。有可能吗？

我看到了此线程：如何将外部自定义URL添加到WooCommerce Endpoint

但是，我在buddypress上的页面无法使用，因为它不能写在ulr中，也不能用写(可能只是我的知识(：

add_filter ( 'woocommerce_account_menu_items', 'add_menu_item_to_tabbed_my_account' );
function add_menu_item_to_tabbed_my_account( $menu_links ){
    // we will hook "anyuniquetext123" later
    $new_item = array( 'anyuniquetext123' => __('Candidate Dashboard') );
    // or in case you need 2 links
    // $new_item = array( 'link1' => 'Link 1', 'link2' => 'Link 2' );
    // array_slice() is good when you want to add an element between the other ones
    $menu_links = array_slice( $menu_links, 0, 1, true ) 
    + $new_item 
    + array_slice( $menu_links, 1, NULL, true );
    return $menu_links;
}
add_filter( 'woocommerce_get_endpoint_url', 'custom_endpoint_url', 10, 4 );
function custom_endpoint_url( $url, $endpoint, $value, $permalink ){
    if( $endpoint === 'anyuniquetext123' ) {
        // BuddyPress My page link → <?php echo bp_loggedin_user_domain(); ?>
        // I want to make a link of this
        // ok, here is the place for your custom URL, it could be external
        $url = **'http://alatta.org.ye/candidate-dashboard/';**
    }
    return $url;
}

实际上，除了主要问题外，当我更改名称时，我会遇到错误。

[更改]'**候选仪表板**' - ＆gt;'AAA'

[错误消息]语法错误，意外''(t_string(

Answer 1

我用以下内容安全地解决了它：

add_filter ( 'woocommerce_account_menu_items', 'add_menu_item_to_tabbed_my_account' );
function add_menu_item_to_tabbed_my_account( $menu_links ){
    $new_item = array( 'anyuniquetext123' => __('Candidate Dashboard') );
    $menu_links = array_slice( $menu_links, 0, 1, true ) 
    + $new_item 
    + array_slice( $menu_links, 1, NULL, true );
 
    return $menu_links;
}
 
add_filter( 'woocommerce_get_endpoint_url', 'custom_endpoint_url', 10, 4 );
function custom_endpoint_url( $url, $endpoint, $value, $permalink ){
    if( $endpoint === 'anyuniquetext123' ) {
 
        // I only had to describe it normally
        $url =  bp_loggedin_user_domain();
    }
    
    return $url;
}