还有比这更好/更快的代码吗？它显示了作者曾经在

我有一个显示用户发布的类别的查询。例如，如果我有3只猫，名字是新闻、媒体、出版物，而作者只在前两个发布过，它会显示

消息

媒体

有没有比这更好的代码，或者有什么改进，因为我的db真的很大，它会让我感到窒息？非常感谢。

<?php
$author = get_query_var(\'author\');
$categories = $wpdb->get_results("
SELECT DISTINCT(terms.term_id) as ID, terms.name
    FROM $wpdb->posts as posts
    LEFT JOIN $wpdb->term_relationships as relationships ON posts.ID = relationships.object_ID
    LEFT JOIN $wpdb->term_taxonomy as tax ON relationships.term_taxonomy_id = tax.term_taxonomy_id
    LEFT JOIN $wpdb->terms as terms ON tax.term_id = terms.term_id
WHERE posts.post_author = \'{$post->post_author}\' ");
?>

<ul>
    <?php foreach($categories as $category) : ?>
<?php if ( ($category->ID == \'3\')   || ($category->ID == \'4\')  || ($category->ID == \'5\')) { ?>
    <li>
        <a href="<?php echo get_category_link( $category->ID ); ?>/?author_name=<?php echo $curuser->user_login; ?>" title="<?php echo $category->name ?>">
            <?php echo $category->name; ?>
        </a>
    </li>
<?php } ?>
    <?php endforeach; ?>
</ul>

SELECT DISTINCT(terms.term_id) as ID, terms.name FROM $wpdb->posts as posts LEFT JOIN $wpdb->term_relationships as relationships ON posts.ID = relationships.object_ID LEFT JOIN $wpdb->term_taxonomy as tax ON (relationships.term_taxonomy_id = tax.term_taxonomy_id AND tax.taxonomy = "category") LEFT JOIN $wpdb->terms as terms ON tax.term_id = terms.term_id WHERE posts.post_author = {$post->post_author}

SELECT terms.term_id AS ID, terms.name FROM $wpdb->posts AS posts LEFT JOIN $wpdb->term_relationships AS relationships ON posts.ID = relationships.object_ID LEFT JOIN $wpdb->term_taxonomy AS tax ON ( relationships.term_taxonomy_id = tax.term_taxonomy_id AND tax.taxonomy = "category" ) LEFT JOIN $wpdb->terms AS terms ON tax.term_id = terms.term_id WHERE posts.post_author = {$post->post_author} AND terms.term_id IS NOT NULL GROUP BY terms.term_id

小码农CODE

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