Question

我可以从WordPress数据库中获取这样的特定类别的链接（为了清楚起见，我在SQL部分添加了换行符）：

$category1 = 'stuff';
$category2 = 'other_stuff';

$q = 'select * from wp_links l
    inner join wp_term_relationships r on l.link_id = r.object_id
    inner join wp_term_taxonomy using (term_taxonomy_id)
    inner join wp_terms using (term_id)
    where taxonomy = "link_category"
    and name = '.$category1;

$results = $wpdb->get_results($q);

如何检索 $category1和$category2中的链接（我的意思是两个类别，而不是 < / em> category）？

Answer 1

你试过OR？

   $q = 'select * from (select * from wp_links l
        inner join wp_term_relationships r on l.link_id = r.object_id
        inner join wp_term_taxonomy using (term_taxonomy_id)
        inner join wp_terms using (term_id)
        where taxonomy = "link_category"
        and name = '.$category1.') as t1 join (select * from wp_links l
        inner join wp_term_relationships r on l.link_id = r.object_id
        inner join wp_term_taxonomy using (term_taxonomy_id)
        inner join wp_terms using (term_id)
        where taxonomy = "link_category"
        and name = '.$category2 .') as t2
        on t1.term_id=t2.term_id;

我不认为这是最好的解决方案，但是你没有给我更多关于你的结构的详细信息，表格你有什么

Answer 2

假设您正在搜索的name字段位于wp_terms表中，则有两个主要选项。

第一个只查找具有第一个类别的实体，然后使用另一个连接查找那些也具有第二个类别的实体。这个involes在两个不同的连接中使用相同的表，因此使用别名并避免使用using关键字。

select * from wp_links           l
inner join wp_term_relationships r  on l.link_id          = r.object_id
inner join wp_term_taxonomy      t  on r.term_taxonomy_id = t.term_taxonomy_id
inner join wp_terms              c1 on t.term_id          = c1.term_id
inner join wp_terms              c2 on t.term_id          = c2.term_id
where taxonomy = "link_category"
and c1.name = 'stuff'
and c2.name = 'other_stuff'

替代方案可扩展到不仅仅是两个类别，但涉及子查询......

select
  l.*
from
(
  select l.id from wp_links        l
  inner join wp_term_relationships r on l.link_id          = r.object_id
  inner join wp_term_taxonomy      t on r.term_taxonomy_id = t.term_taxonomy_id
  inner join wp_terms              c on t.term_id          = c.term_id
  where taxonomy = "link_category"
  and c.name IN ('stuff', 'other_stuff')
  group by l.id
  having count(distinct c.name) = 2
)
  subquery
inner join wp_links l ON subquery.id = l.id

内部查询查找具有一个类别或另一个类别的所有内容，但是having子句仅允许通过列表中包含两个类别的那些。（换句话说，两者都有。） [它还假定wp_links表作为id列用作唯一标识符。]

多个类别中的wp_links的SQL查询

2 个答案: