Question

我正在创建以下wordpress查询哪个输出正在计算记录。但是所有帖子都有一个名为betting_odds的meta_key，其中meta_value类似于2.05,3.30

如何将此meta_key的平均值添加到以下查询中？

SELECT count(DISTINCT $wpdb->postmeta.`post_id`) 
  FROM $wpdb->posts
       LEFT JOIN $wpdb->postmeta ON ($wpdb->posts.ID = $wpdb->postmeta.post_id)
       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) 
 WHERE $wpdb->postmeta.meta_key       = 'betting_status'
    AND $wpdb->postmeta.meta_value    != 'Ikke afgjort'
    AND $wpdb->posts.post_status      = 'publish'
    AND $wpdb->term_taxonomy.taxonomy = 'category'
    AND $wpdb->term_taxonomy.term_id  = 106

Answer 1

从wp_postmeta等键/值存储区检索数据时，这是一个常见的挑战。

诀窍是将post_id和meta_key放入ON子句中，就像这样。

（在WordPress查询中使用表别名（在本例中为posts，stat，odds等）也很有帮助，因为它们可以获得更多功能可读的。

SELECT count(DISTINCT posts.ID), AVG(odds.meta_value)
  FROM $wpdb->posts posts
  LEFT JOIN $wpdb->postmeta stat
                 ON posts.ID = stat.post_id
                AND stat.meta_key = 'betting_status'
  LEFT JOIN $wpdb->postmeta odds
                 ON posts.ID = odds.post_id
                AND odds.meta_key = 'betting_odds'
  LEFT JOIN $wpdb->term_relationships tr ON posts.ID = tr.object_id
  LEFT JOIN $wpdb->term_taxonomy t ON tr.term_taxonomy_id = t.term_taxonomy_id 
 WHERE stat.meta_value    != 'Ikke afgjort'
   AND posts.post_status      = 'publish'
   AND t.taxonomy = 'category'
   AND t.term_id  = 106

你是否看到我们如何LEFT JOIN两次postmeta表，我们需要的每个元值一次？您是否了解meta_key匹配条件如何进入ON子句而不是WHERE子句？这是为帖子提取元值的模式。

从wp的元值中获取平均数

1 个答案: