在MySQL中使用in IN WHERE子句执行内连接时避免全表扫描

Question

在MySQL中使用in IN WHERE子句执行内部连接时，如何避免全表扫描？例如：

explain SELECT
-> COUNT(DISTINCT(n.nid))
-> FROM node n
-> INNER JOIN term_node tn ON n.nid = tn.nid
-> INNER JOIN content_type_article ca ON n.nid = ca.nid
-> WHERE tn.tid IN (67,100)
-> ;
+----+-------------+-------+--------+----------------------------------+---------+---------+----------------------+-------+--------------------------+
| id | select_type | table | type   | possible_keys                    | key     | key_len | ref                  | rows  | Extra                    |
+----+-------------+-------+--------+----------------------------------+---------+---------+----------------------+-------+--------------------------+
|  1 | SIMPLE      | tn    | ALL    | PRIMARY,nid                      | NULL    | NULL    | NULL                 | 42180 | Using where              |
|  1 | SIMPLE      | ca    | ref    | nid,field_article_date_nid_index | nid     | 4       | drupal_mm_qas.tn.nid |     1 | Using index              |
|  1 | SIMPLE      | n     | eq_ref | PRIMARY                          | PRIMARY | 4       | drupal_mm_qas.ca.nid |     1 | Using where; Using index |
+----+-------------+-------+--------+----------------------------------+---------+---------+----------------------+-------+--------------------------+
3 rows in set (0.00 sec)

Answer 1

您似乎是在按mysql认为不够有选择性的列进行筛选。当筛选器的基数太低(即，该筛选器的不同行的数量很低)时，mysql认为，在大多数情况下，fts会更快。

为了确认，请显示SELECT COUNT(DISTINCT tn.tid) FROM term_node tn和SELECT COUNT(*) FROM term_node tn的结果