Question

简单的问题。我有以下自定义查询。如何通过“echo？”生成数组的html列表

$q_result = $wpdb->get_col("SELECT DISTINCT {$wpdb->terms}.name FROM {$wpdb->terms}
        INNER JOIN {$wpdb->term_taxonomy} ON {$wpdb->term_taxonomy}.term_id = {$wpdb->terms}.term_id
        INNER JOIN {$wpdb->term_relationships} ON {$wpdb->term_taxonomy}.term_taxonomy_id = {$wpdb->term_relationships}.term_taxonomy_id
        WHERE {$wpdb->term_taxonomy}.taxonomy = 'series' AND {$wpdb->term_relationships}.object_id IN (
        SELECT object_id FROM {$wpdb->term_relationships}
        INNER JOIN {$wpdb->term_taxonomy} ON {$wpdb->term_taxonomy}.term_taxonomy_id = {$wpdb->term_relationships}.term_taxonomy_id
        WHERE {$wpdb->term_taxonomy}.taxonomy = 'media_type' AND {$wpdb->term_taxonomy}.term_id = '16'
        ORDER BY {$wpdb->terms}.term_order DESC
        ) ORDER BY {$wpdb->terms}.term_order ASC");

Answer 1

您可以使用foreach -

迭代数组

foreach( $q_result as $result ){
   echo "<a href='/series/".$result."/>".$result."</a>";
}

如果您希望它也显示HTML，则必须回显正确的HTML标记。以上将为该术语的链接工作 - 格式为TAXONOMY / TERM_SLUG但是您只返回名称，您需要使用get_results()获取名称和slug才能正确执行。

使用表别名重新格式化示例，返回术语和slug：

$sql = <<<SQL
    SELECT DISTINCT t.name, t.slug 
    FROM {$wpdb->terms} t
    INNER JOIN {$wpdb->term_taxonomy} tt 
        ON tt.term_id = t.term_id and tt.taxonomy = 'series'
    INNER JOIN {$wpdb->term_relationships} tr 
        ON tt.term_taxonomy_id = tr.term_taxonomy_id
    WHERE tr.object_id IN (
        SELECT object_id 
        FROM {$wpdb->term_relationships} tr2
        INNER JOIN {$wpdb->term_taxonomy} tt2 
            ON tt2.term_taxonomy_id = tr2.term_taxonomy_id AND tt2.taxonomy = 'media_type' AND tt2.term_id = '16'
        ORDER BY {$wpdb->terms}.term_order DESC
    ) 
    ORDER BY {$wpdb->terms}.term_order ASC
SQL;

$terms = $wpdb->get_results( $sql );
for ( $terms as $term ){
    echo "<a href='/series/{$term->slug}/'>{$term->name}</a><br/>";
}

自定义查询结果生成代码

1 个答案: