[英]Avoiding full table scan when performing inner joins in MySQL using IN in WHERE clause
在WHERE子句中使用IN在MySQL中执行内部联接时,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)
看来您正在筛选一列mysql识别为选择性不够的列。 当一个过滤器的基数太低(即该过滤器的不同行数很低)时,mysql认为,在大多数情况下,准确地说,英尺数会更快。
为了确认,请显示SELECT COUNT(DISTINCT tn.tid) FROM term_node tn
的SELECT COUNT(DISTINCT tn.tid) FROM term_node tn
SELECT COUNT(*) FROM term_node tn
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