Question

我在php语言之外查询wordpress mysql数据库，我不知道如何解决以下问题：每个帖子都有一些有价值的分类法。有一个名为“calificaciones＆＃39;我想要过滤的。这是我目前的查询：

SELECT wp_posts.post_title, IF (wp_term_taxonomy.taxonomy = 'calificaciones', wp_terms.slug, 'no') as calificacion
    FROM wp_posts
    JOIN wp_term_relationships ON (wp_term_relationships.object_id = wp_posts.ID)
    JOIN wp_term_taxonomy ON (wp_term_taxonomy.term_taxonomy_id = wp_term_relationships.term_taxonomy_id)
    JOIN wp_terms ON (wp_terms.term_id = wp_term_taxonomy.term_id)

    AND wp_posts.post_type = 'ultimas-noticias'
    #AND wp_term_taxonomy.taxonomy = 'calificaciones'

此查询的问题在于我会多次获得post_title（因为每个帖子都有其他分类法）。如果帖子具有该分类值，我想获得帖子标题和标志是 - 否。

Answer 1

您可以使用calificacion列的子查询，而不是在同一查询中加入分类法。

SELECT wp_posts.post_title, CASE WHEN EXISTS((SELECT *
    FROM wp_term_relationships
    JOIN wp_term_taxonomy ON (wp_term_taxonomy.term_taxonomy_id = wp_term_relationships.term_taxonomy_id)
    JOIN wp_terms ON (wp_terms.term_id = wp_term_taxonomy.term_id)
    WHERE (wp_term_relationships.object_id = wp_posts.ID)
    AND wp_term_taxonomy.taxonomy = 'calificaciones'
    )) THEN 'YES' ELSE 'NO' END as calificacion
FROM wp_posts
WHERE wp_posts.post_type = 'ultimas-noticias'

Answer 2

您可以使用select distinct

dat.1 #and the output will be all ASCII charaters
[1] "Work, Awareness, Potential"

过滤没有特定值的行

2 个答案: