Question

我试图创建一个自定义的mysql查询，它计算了两件事。

首先，它应该计算meta_key betting_status＆＃39;的类别中的帖子数量。值不等于＆＃34;待定＆＃34;

然后它应该计算meta_key betting_status＆＃39;的类别中的帖子数量。值等于是。

到目前为止，我已经完成了这项工作，但它根本没有输出任何内容：

$wombo_query = "SELECT count(DISTINCT $wpdb->postmeta.post_id) FROM $wpdb->posts
LEFT JOIN $wpdb->term_relationships ON
($wpdb->posts.ID = $wpdb->term_relationships.object_id)
LEFT JOIN $wpdb->term_taxonomy ON
($wpdb->term_relationships.term_taxonomy_id = $wpdb->term_taxonomy.term_taxonomy_id) AND
WHERE $wpdb->postmeta.meta_key = 'betting_status'
AND $wpdb->posts.meta_value = 'yes'
WHERE $wpdb->posts.post_status = 'publish'
AND $wpdb->term_taxonomy.taxonomy = 'category'
AND $wpdb->term_taxonomy.term_id = 106
";

Answer 1

试试这个。

SELECT count(DISTINCT wp_postmeta.`post_id`) 
  FROM wp_posts
       LEFT JOIN wp_postmeta ON (wp_posts.ID = wp_postmeta.post_id)
       LEFT JOIN wp_term_relationships ON (wp_posts.ID = wp_term_relationships.object_id)
       LEFT JOIN wp_term_taxonomy ON (wp_term_relationships.term_taxonomy_id = wp_term_taxonomy.term_taxonomy_id) 
 WHERE wp_postmeta.meta_key       = 'betting_status'
    AND wp_postmeta.meta_value    = 'yes'
    AND wp_posts.post_status      = 'publish'
    AND wp_term_taxonomy.taxonomy = 'category'
    AND wp_term_taxonomy.term_id  = 106;

由于我的数据库中没有与你的数据库相同的值，因此我返回0行，但它确实执行了。

干杯！ = C =

从meta_key等于的类别中获取帖子数量？

1 个答案: