1

目前，我正在执行以下操作以获取所有作者的列表：

$authors = wp_list_authors('html=0&style=none&echo=0&exclude_admin=1&optioncount=0&show_fullname=1&hide_empty=1&orderby=name&order=ASC'); 

$authors_array = explode(',', $authors);

for ($j = 0; $j < count($authors_array); $j++)
{
    echo '<li id="">'.$authors_array[$j].'</li>';
}

我怎样才能得到用户的ID？

我到处寻找，我想不出一种方法来获取所有作者和另一位元数据。

4

2 回答 2

4

我发现的另一种方法是执行以下操作（发誓我之前找不到这样的东西）：

$authors = get_users('role=author&orderby=display_name&order=ASC');

foreach ($authors as $author) {
    if (count_user_posts($author->ID) > 0) {
       echo '<li id="' . $author->ID . '">' . $author->display_name . '</li>';
    }
}
于 2012-08-13T14:17:26.747 回答
1

您可以覆盖functions.php 中的函数，也可以在原始函数的基础上创建一个新函数。查看代码很容易将 $author_id 包含在 foreach 循环中。

原始代码在这里

试试这个并确保将新参数（includeauthorid）添加到函数调用中......

$authors = wp_list_authors('html=0&style=none&echo=0&exclude_admin=1&optioncount=0&show_fullname=1&hide_empty=1&orderby=name&order=ASC&includeauthorid=1');

新功能 - ps 我不建议更改原始功能。只需将其放在functions.php 中即可。

function wp_list_authors($args = '') {
global $wpdb;

$defaults = array(
    'orderby' => 'name', 'order' => 'ASC', 'number' => '',
    'optioncount' => false, 'exclude_admin' => true,
    'show_fullname' => false, 'hide_empty' => true,
    'feed' => '', 'feed_image' => '', 'feed_type' => '', 'echo' => true,
    'style' => 'list', 'html' => true,
    'includeauthorid' => false 
);

$args = wp_parse_args( $args, $defaults );
extract( $args, EXTR_SKIP );

$return = '';

$query_args = wp_array_slice_assoc( $args, array( 'orderby', 'order', 'number' ) );
$query_args['fields'] = 'ids';
$authors = get_users( $query_args );

$author_count = array();
foreach ( (array) $wpdb->get_results("SELECT DISTINCT post_author, COUNT(ID) AS count FROM $wpdb->posts WHERE post_type = 'post' AND " . get_private_posts_cap_sql( 'post' ) . " GROUP BY post_author") as $row )
    $author_count[$row->post_author] = $row->count;

foreach ( $authors as $author_id ) {
    $author = get_userdata( $author_id );

    if ( $exclude_admin && 'admin' == $author->display_name )
        continue;

    $posts = isset( $author_count[$author->ID] ) ? $author_count[$author->ID] : 0;

    if ( !$posts && $hide_empty )
        continue;

    $link = '';

    if ( $show_fullname && $author->first_name && $author->last_name )
        $name = "$author->first_name $author->last_name";
    else
        $name = $author->display_name;

    if( $includeauthorid)
        $name .= ' ('. $author_id .')';

    if ( !$html ) {
        $return .= $name . ', ';

        continue; // No need to go further to process HTML.
    }

    if ( 'list' == $style ) {
        $return .= '<li>';
    }

    $link = '<a href="' . get_author_posts_url( $author->ID, $author->user_nicename ) . '" title="' . esc_attr( sprintf(__("Posts by %s"), $author->display_name) ) . '">' . $name . '</a>';

    if ( !empty( $feed_image ) || !empty( $feed ) ) {
        $link .= ' ';
        if ( empty( $feed_image ) ) {
            $link .= '(';
        }

        $link .= '<a href="' . get_author_feed_link( $author->ID ) . '"';

        $alt = $title = '';
        if ( !empty( $feed ) ) {
            $title = ' title="' . esc_attr( $feed ) . '"';
            $alt = ' alt="' . esc_attr( $feed ) . '"';
            $name = $feed;
            $link .= $title;
        }

        $link .= '>';

        if ( !empty( $feed_image ) )
            $link .= '<img src="' . esc_url( $feed_image ) . '" style="border: none;"' . $alt . $title . ' />';
        else
            $link .= $name;

        $link .= '</a>';

        if ( empty( $feed_image ) )
            $link .= ')';
    }

    if ( $optioncount )
        $link .= ' ('. $posts . ')';

    $return .= $link;
    $return .= ( 'list' == $style ) ? '</li>' : ', ';
}

$return = rtrim($return, ', ');

if ( !$echo )
    return $return;

echo $return;

}

于 2012-08-13T12:51:54.663 回答