我正在尝试多次执行以下查询：
SELECT st2.stop_id AS to_stop_id,
TIME_TO_SEC(
ADDTIME(TIMEDIFF(MIN(st1.time), %time),
TIMEDIFF(st2.time, st2.time))) AS duration
FROM stop_times st1,
stop_times st2,
trips tr,
calendar cal
WHERE tr.service_id = cal.service_id
AND tr.trip_id = st1.trip_id
AND st1.trip_id = st2.trip_id
AND st1.stop_id = %sid
AND st1.stop_seq +1 = st2.stop_seq
AND st1.time > %time
AND DATE(NOW()) BETWEEN cal.start_date AND
cal.end_date
GROUP BY st2.stop_id
然而，它运行得非常慢。我索引了以下属性：
+------------+------------+------------+--------------+-------------+-----------+-------------+----------+--------+------+------------+---------+---------------+
| Table | Non_unique | Key_name | Seq_in_index | Column_name | Collation | Cardinality | Sub_part | Packed | Null | Index_type | Comment | Index_comment |
+------------+------------+------------+--------------+-------------+-----------+-------------+----------+--------+------+------------+---------+---------------+
| stop_times | 0 | st_id | 1 | st_id | A | 11431583 | NULL | NULL | | BTREE | | |
| stop_times | 1 | fk_tid_s | 1 | trip_id | A | 1039234 | NULL | NULL | YES | BTREE | | |
| stop_times | 1 | st_per_sid | 1 | stop_id | A | 33135 | NULL | NULL | YES | BTREE | | |
| calendar | 0 | PRIMARY | 1 | service_id | A | 5206 | NULL | NULL | | BTREE | | |
| calendar | 0 | PRIMARY | 1 | service_id | A | 5206 | NULL | NULL | | BTREE | | |
| trips | 0 | PRIMARY | 1 | trip_id | A | 449489 | NULL | NULL | | BTREE | | |
| trips | 1 | fk_rid | 1 | route_id | A | 1937 | NULL | NULL | YES | BTREE | | |
| trips | 1 | fk_sid | 1 | service_id | A | 7749 | NULL | NULL | YES | BTREE | | |
+------------+------------+------------+--------------+-------------+-----------+-------------+----------+--------+------+------------+---------+---------------+
（由于某些原因，st_id并未显示为PRIMARY KEY，但它是，我不知道它是否重要，但以防万一..）
我在这个查询上运行了SQL EXPLAIN，它给了我以下答案：
+------+-------------+-------+--------+-------------------------------------------------+---------------------+---------+------------------------------+------+---------------------------------------------------------------------+
| id | select_type | table | type | possible_keys | key | key_len | ref | rows | Extra |
+------+-------------+-------+--------+-------------------------------------------------+---------------------+---------+------------------------------+------+---------------------------------------------------------------------+
| 1 | SIMPLE | st1 | range | comp_uniq_st_seq,st_per_sid,comp_uniq_stid_time | comp_uniq_stid_time | 9 | NULL | 1396 | Using index condition; Using where; Using temporary; Using filesort |
| 1 | SIMPLE | tr | eq_ref | PRIMARY,fk_sid | PRIMARY | 8 | reseau_ratp.st1.trip_id | 1 | Using where |
| 1 | SIMPLE | cal | eq_ref | PRIMARY,comp_sid_date_en,comp_sid_date_st | PRIMARY | 4 | reseau_ratp.tr.service_id | 1 | Using where |
| 1 | SIMPLE | st2 | ref | comp_uniq_st_seq | comp_uniq_st_seq | 14 | reseau_ratp.st1.trip_id,func | 1 | Using index condition |
+------+-------------+-------+--------+-------------------------------------------------+---------------------+---------+------------------------------+------+---------------------------------------------------------------------+
我该怎么做才能让这个查询更快地运行？
编辑： 使用请求的语法进行查询：
SELECT st2.stop_id AS to_stop_id,
TIME_TO_SEC(
ADDTIME(TIMEDIFF(MIN(st1.time), %time),
TIMEDIFF(st2.time, st2.time))) AS duration
FROM stop_times st1
INNER JOIN stop_times st2
ON st1.trip_id = st2.trip_id AND st1.stop_seq + 1 = st2.stop_seq
INNER JOIN trips tr
ON tr.trip_id = st1.trip_id
INNER JOIN calendar cal
ON tr.service_id = cal.service_id
WHERE st1.stop_id = %sid
AND st1.time > %time
AND cal.start_date <= NOW()
AND cal.end_date >= NOW()
GROUP BY st2.stop_id
此处SHOW CREATE TABLE stop_times：
CREATE TABLE `stop_times` (
`trip_id` bigint(10) unsigned DEFAULT NULL,
`stop_id` int(10) DEFAULT NULL,
`time` time DEFAULT NULL,
`stop_seq` int(10) unsigned DEFAULT NULL,
UNIQUE KEY `comp_uniq_st_seq` (`trip_id`,`stop_seq`),
KEY `comp_uniq_stid_time` (`stop_id`,`time`),
CONSTRAINT `fk_sid_s` FOREIGN KEY (`stop_id`) REFERENCES `stops` (`stop_id`),
CONSTRAINT `fk_tid_s` FOREIGN KEY (`trip_id`) REFERENCES `trips` (`trip_id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8
calendar：
CREATE TABLE `calendar` (
`service_id` int(10) unsigned NOT NULL,
`start_date` date DEFAULT NULL,
`end_date` date DEFAULT NULL,
PRIMARY KEY (`service_id`),
KEY `comp_sid_date_en` (`service_id`,`end_date`),
KEY `comp_sid_date_st` (`service_id`,`start_date`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8
对于trips：
CREATE TABLE `trips` (
`trip_id` bigint(10) unsigned NOT NULL DEFAULT '0',
`route_id` int(10) unsigned DEFAULT NULL,
`service_id` int(10) unsigned DEFAULT NULL,
`trip_headsign` varchar(15) DEFAULT NULL,
`trip_short_name` varchar(15) DEFAULT NULL,
`direction_id` tinyint(1) DEFAULT NULL,
PRIMARY KEY (`trip_id`),
KEY `fk_rid` (`route_id`),
KEY `fk_sid` (`service_id`),
CONSTRAINT `fk_rid` FOREIGN KEY (`route_id`) REFERENCES `routes` (`route_id`),
CONSTRAINT `fk_sid` FOREIGN KEY (`service_id`) REFERENCES `calendar` (`service_id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8
答案 0 :(得分：1)
st1需要此综合索引：INDEX(stop_id, time)
请使用JOIN ... ON语法。
请提供SHOW CREATE TABLE。
这是Cookbook on creating INDEXes from a SELECT。
（编辑）
Calendar处理起来比较棘手，没有好的＆＃34;指数。这些可能有所帮助：
INDEX(service_id, start_time)
INDEX(service_id, end_time)
加上，将AND DATE(NOW()) BETWEEN cal.start_date AND cal.end_date改为
AND cal.start_date <= NOW()
AND cal.end_time >= NOW()
（编辑2）
只要可行，请说NOT NULL。这在stop_times没有PRIMARY KEY时尤为重要。将UNIQUE KEY comp_uniq_st_seq（trip_id，stop_seq）中的两列更改为NOT NULL ，将其转换为PRIMARY KEY (trip_id, stop_seq)。这将使得PK的性能优势与数据＆＃34;踢进去。
现在我看到了CREATE TABLE的日历，而service_id是PRIMARY KEY，我建议的两个索引可能没用。 （同样，这涉及＆＃34;聚类＆＃34;。）
我的Cookbook for building indexes可能会派上用场。