MySQL Cookbook 学习笔记-03

1、INNER JOIN关联查询
2、outer join（LEFT JOIN 与 RIGHT JOIN）
3、自连接
4、主从表查询
5、在分组内查找某列最大或最小的一行
6、计算小组积分榜
7、计算连续行的差
8、计算“累计和”与运行时平均值
9、使用 JOIN 控制查询结果的顺序
10、通过 UNION 合并查询结果集

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1、INNER JOIN关联查询

（Recipe 12.1. Finding Rows in One Table That Match Rows in Another）

CREATE TABLE artist(  a_id  INT UNSIGNED NOT NULL AUTO_INCREMENT, # artist ID  name  VARCHAR(30) NOT NULL,                 # artist name  PRIMARY KEY (a_id),  UNIQUE (name));CREATE TABLE painting(  a_id  INT UNSIGNED NOT NULL,                # artist ID  p_id  INT UNSIGNED NOT NULL AUTO_INCREMENT, # painting ID  title VARCHAR(100) NOT NULL,                # title of painting  state VARCHAR(2) NOT NULL,                  # state where purchased  price INT UNSIGNED,                         # purchase price (dollars)  INDEX (a_id),  PRIMARY KEY (p_id));

表的数据：

mysql> SELECT * FROM artist ORDER BY a_id;+------+----------+| a_id | name     |+------+----------+|    1 | Da Vinci ||    2 | Monet    ||    3 | Van Gogh ||    4 | Picasso  ||    5 | Renoir   |+------+----------+

mysql> SELECT * FROM painting ORDER BY a_id, p_id;+------+------+-------------------+-------+-------+| a_id | p_id | title             | state | price |+------+------+-------------------+-------+-------+|    1 |    1 | The Last Supper   | IN    |    34 ||    1 |    2 | The Mona Lisa     | MI    |    87 ||    3 |    3 | Starry Night      | KY    |    48 ||    3 |    4 | The Potato Eaters | KY    |    67 ||    3 |    5 | The Rocks         | IA    |    33 ||    5 |    6 | Les Deux Soeurs   | NE    |    64 |+------+------+-------------------+-------+-------+

方案一：通过 where 语句关联查询

mysql> SELECT * FROM artist, painting    -> WHERE artist.a_id = painting.a_id;

方案二：通过 INNER JOIN .... ON 关联查询（等价方案一）

mysql> SELECT * FROM artist INNER JOIN painting    -> ON artist.a_id = painting.a_id;+------+----------+------+------+-------------------+-------+-------+| a_id | name     | a_id | p_id | title             | state | price |+------+----------+------+------+-------------------+-------+-------+|    1 | Da Vinci |    1 |    1 | The Last Supper   | IN    |    34 ||    1 | Da Vinci |    1 |    2 | The Mona Lisa     | MI    |    87 ||    3 | Van Gogh |    3 |    3 | Starry Night      | KY    |    48 ||    3 | Van Gogh |    3 |    4 | The Potato Eaters | KY    |    67 ||    3 | Van Gogh |    3 |    5 | The Rocks         | IA    |    33 ||    5 | Renoir   |    5 |    6 | Les Deux Soeurs   | NE    |    64 |+------+----------+------+------+-------------------+-------+-------+

方案三：通过 INNER JOIN ... USING 关联查询（关联两表的列必须相同）

mysql> SELECT * FROM artist INNER JOIN painting    -> USING(a_id);+------+----------+------+-------------------+-------+-------+| a_id | name     | p_id | title             | state | price |+------+----------+------+-------------------+-------+-------+|    1 | Da Vinci |    1 | The Last Supper   | IN    |    34 ||    1 | Da Vinci |    2 | The Mona Lisa     | MI    |    87 ||    3 | Van Gogh |    3 | Starry Night      | KY    |    48 ||    3 | Van Gogh |    4 | The Potato Eaters | KY    |    67 ||    3 | Van Gogh |    5 | The Rocks         | IA    |    33 ||    5 | Renoir   |    6 | Les Deux Soeurs   | NE    |    64 |+------+----------+------+-------------------+-------+-------+

推荐用法：JOIN 语句来关联表，where 子句作为筛选条件

mysql> SELECT artist.name, painting.title, painting.state, painting.price    -> FROM artist INNER JOIN painting    -> ON artist.a_id = painting.a_id    -> WHERE painting.state = 'KY';+----------+-------------------+-------+-------+| name     | title             | state | price |+----------+-------------------+-------+-------+| Van Gogh | Starry Night      | KY    |    48 || Van Gogh | The Potato Eaters | KY    |    67 |+----------+-------------------+-------+-------+

三表关联示例：

mysql> SELECT artist.name, painting.title, states.name, painting.price    -> FROM artist INNER JOIN painting INNER JOIN states    -> ON artist.a_id = painting.a_id AND painting.state = states.abbrev;+----------+-------------------+----------+-------+| name     | title             | name     | price |+----------+-------------------+----------+-------+| Da Vinci | The Last Supper   | Indiana  |    34 || Da Vinci | The Mona Lisa     | Michigan |    87 || Van Gogh | Starry Night      | Kentucky |    48 || Van Gogh | The Potato Eaters | Kentucky |    67 || Van Gogh | The Rocks         | Iowa     |    33 || Renoir   | Les Deux Soeurs   | Nebraska |    64 |+----------+-------------------+----------+-------+

关联查询和聚合函数——示例

mysql> SELECT artist.name,    -> COUNT(*) AS 'number of paintings',    -> SUM(painting.price) AS 'total price',    -> AVG(painting.price) AS 'average price'    -> FROM artist INNER JOIN painting ON artist.a_id = painting.a_id    -> GROUP BY artist.name;+----------+---------------------+-------------+---------------+| name     | number of paintings | total price | average price |+----------+---------------------+-------------+---------------+| Da Vinci |                   2 |         121 |       60.5000 || Renoir   |                   1 |          64 |       64.0000 || Van Gogh |                   3 |         148 |       49.3333 |+----------+---------------------+-------------+---------------+

特别注意：为了提高联合查询的效率，一般将联合的列（如：a_id）设定为索引！

TABLE artist，a_id 为主键（强制索引）

TABLE painting，a_id 直接设定为索引

2、outer join（LEFT JOIN 与 RIGHT JOIN）

查询在表painting 中没有作品的作者：

mysql> select artist.a_id, artist.name from artist    -> left join painting on artist.a_id = painting.a_id    -> where painting.a_id is null;+------+---------+| a_id | name    |+------+---------+|    2 | Monet   ||    4 | Picasso |+------+---------+

其它方法（非 JOIN）

mysql> select artist.a_id, artist.name from artist    -> where artist.a_id not in (    -> select painting.a_id from painting    -> );+------+---------+| a_id | name    |+------+---------+|    2 | Monet   ||    4 | Picasso |+------+---------+

LEFT JOIN 和 RIGHT JOIN 的相互转换：

mysql> SELECT artist.name,    -> IF(COUNT(painting.a_id)>0,'yes','no') AS 'in collection'    -> FROM artist LEFT JOIN painting ON artist.a_id = painting.a_id    -> GROUP BY artist.name;+----------+---------------+| name     | in collection |+----------+---------------+| Da Vinci | yes           || Monet    | no            || Picasso  | no            || Renoir   | yes           || Van Gogh | yes           |+----------+---------------+

mysql> SELECT artist.name,    -> IF(COUNT(painting.a_id)>0,'yes','no') AS 'in collection'    -> FROM painting RIGHT JOIN artist ON artist.a_id = painting.a_id    -> GROUP BY artist.name;+----------+---------------+| name     | in collection |+----------+---------------+| Da Vinci | yes           || Monet    | no            || Picasso  | no            || Renoir   | yes           || Van Gogh | yes           |+----------+---------------+

3、自连接

任务：查找作品为“The Potato Eaters”的作者的所有作品。

mysql> select p2.title    -> from painting as p1 inner join painting as p2    -> on p1.a_id = p2.a_id    -> where p1.title = 'The Potato Eaters';+-------------------+| title             |+-------------------+| Starry Night      || The Potato Eaters || The Rocks         |+-------------------+

替换方法（嵌套查询）

mysql> select title from painting    -> where painting.a_id = (    -> select a_id from painting where title = 'The Potato Eaters'    -> );+-------------------+| title             |+-------------------+| Starry Night      || The Potato Eaters || The Rocks         |+-------------------+

4、主从表查询

a、只查询从表中有数据的“主-从”关联查询，用 INNER  JOIN

b、查询主表中所有数据，从表可以没有，用 LEFT  JOIN

a——示例

mysql> select artist.name, painting.title    -> from artist INNER JOIN painting ON artist.a_id = painting.a_id    -> order by name, title;+----------+-------------------+| name     | title             |+----------+-------------------+| Da Vinci | The Last Supper   || Da Vinci | The Mona Lisa     || Renoir   | Les Deux Soeurs   || Van Gogh | Starry Night      || Van Gogh | The Potato Eaters || Van Gogh | The Rocks         |+----------+-------------------+

带聚集函数——示例

mysql> select artist.name as painter, count(painting.a_id) as count    -> from artist INNER JOIN painting ON artist.a_id = painting.a_id    -> group by artist.name;+----------+-------+| painter  | count |+----------+-------+| Da Vinci |     2 || Renoir   |     1 || Van Gogh |     3 |+----------+-------+

b——示例

mysql> select artist.name, painting.title    -> from artist LEFT JOIN painting ON artist.a_id = painting.a_id    -> order by name, title;+----------+-------------------+| name     | title             |+----------+-------------------+| Da Vinci | The Last Supper   || Da Vinci | The Mona Lisa     || Monet    | NULL              || Picasso  | NULL              || Renoir   | Les Deux Soeurs   || Van Gogh | Starry Night      || Van Gogh | The Potato Eaters || Van Gogh | The Rocks         |+----------+-------------------+

带聚集函数——示例

mysql> select a.a_id, a.name, p.a_id, count(p.a_id) as count    -> from artist as a LEFT JOIN painting as p ON a.a_id = p.a_id    -> group by a.name;+------+----------+------+-------+| a_id | name     | a_id | count |+------+----------+------+-------+|    1 | Da Vinci |    1 |     2 ||    2 | Monet    | NULL |     0 ||    4 | Picasso  | NULL |     0 ||    5 | Renoir   |    5 |     1 ||    3 | Van Gogh |    3 |     3 |+------+----------+------+-------+

聚集函数会忽略 NULL 值，count( ) 会将 NULL 值操作返回0！

例如：

mysql> select * from taxpayer;+---------+--------+| name    | id     |+---------+--------+| bernina | 198-48 || bertha  | NULL   || ben     | NULL   || NULL    | 475-83 || baidu   | 111+55 |+---------+--------+mysql> select count(id) from taxpayer    -> where name='ben';+-----------+| count(id) |+-----------+|         0 |+-----------+

其他聚集函数（如，SUM( )，AVG( )）就必须做特殊处理：

mysql> select a.name as painter,    -> count(p.a_id) as 'number of paintings',    -> sum(p.price) as 'total price',    -> avg(p.price) as 'average price'    -> from artist as a LEFT JOIN painting as p ON a.a_id = p.a_id    -> group by a.name;+----------+---------------------+-------------+---------------+| painter  | number of paintings | total price | average price |+----------+---------------------+-------------+---------------+| Da Vinci |                   2 |         121 |       60.5000 || Monet    |                   0 |        NULL |          NULL || Picasso  |                   0 |        NULL |          NULL || Renoir   |                   1 |          64 |       64.0000 || Van Gogh |                   3 |         148 |       49.3333 |+----------+---------------------+-------------+---------------+

这里的 SUM( ) 和 AVG( )函数返回的都为 NULL ，显然不合理，修改如下：

mysql> select a.name as painter,    -> count(p.a_id) as 'number of paintings',    -> IFNULL(sum(p.price),0) as 'total price',    -> IFNULL(avg(p.price),0) as 'average price'    -> from artist as a LEFT JOIN painting as p ON a.a_id = p.a_id    -> group by a.name;+----------+---------------------+-------------+---------------+| painter  | number of paintings | total price | average price |+----------+---------------------+-------------+---------------+| Da Vinci |                   2 |         121 |       60.5000 || Monet    |                   0 |           0 |        0.0000 || Picasso  |                   0 |           0 |        0.0000 || Renoir   |                   1 |          64 |       64.0000 || Van Gogh |                   3 |         148 |       49.3333 |+----------+---------------------+-------------+---------------+

5、在分组内查找某列最大或最小的一行

（Recipe 12.6. Finding Rows Containing Per-Group Minimum or Maximum Values）

查找全表某列最大或最小的行，如下方法合适：

mysql> select a.name, p.title, p.price    -> from artist as a INNER JOIN painting as p    -> ON p.a_id = a.a_id    -> where p.price = (select max(price) from painting);+----------+---------------+-------+| name     | title         | price |+----------+---------------+-------+| Da Vinci | The Mona Lisa |    87 |+----------+---------------+-------+

查找分组内某列最大或最小的行，错误方法（我的）：

mysql> select a.name, p.title, max(p.price)    -> from artist as a INNER JOIN painting as p    -> ON a.a_id = p.a_id    -> group by a.name;+----------+-----------------+--------------+| name     | title           | max(p.price) |+----------+-----------------+--------------+| Da Vinci | The Last Supper |           87 || Renoir   | Les Deux Soeurs |           64 || Van Gogh | Starry Night    |           67 |+----------+-----------------+--------------+

分析原因，看两表连接查询数据：

mysql> select a.name, p.title, p.price    -> from artist as a INNER JOIN painting as p    -> ON a.a_id = p.a_id;+----------+-------------------+-------+| name     | title             | price |+----------+-------------------+-------+| Da Vinci | The Last Supper   |    34 || Da Vinci | The Mona Lisa     |    87 || Renoir   | Les Deux Soeurs   |    64 || Van Gogh | Starry Night      |    48 || Van Gogh | The Potato Eaters |    67 || Van Gogh | The Rocks         |    33 |+----------+-------------------+-------+

总结：虽然我的方法中“name”列和“max(price)”列都是正确的，但是“title”列却是任意的！
正确方法：

a、单独创建临时表，与临时表关联查询

b、同过 FROM 子句连接查询

c、通过自连接连接查询（比较难懂）

a——示例

mysql> create table tmp    -> select a_id, max(price) as max_price from painting group by a_id;    //演示用-STARTmysql> select * from tmp;+------+-----------+| a_id | max_price |+------+-----------+|    1 |        87 ||    3 |        67 ||    5 |        64 |+------+-----------+//演示用-ENDmysql> select a.name, p.title, p.price    -> from artist as a INNER JOIN painting as p INNER JOIN tmp as t    -> ON a.a_id = p.a_id    -> AND p.a_id = t.a_id    -> AND p.price = t.max_price;+----------+-------------------+-------+| name     | title             | price |+----------+-------------------+-------+| Da Vinci | The Mona Lisa     |    87 || Van Gogh | The Potato Eaters |    67 || Renoir   | Les Deux Soeurs   |    64 |+----------+-------------------+-------+

b——示例

mysql> select a.name, p.title, p.price    -> from artist as a INNER JOIN painting as p    -> INNER JOIN (select a_id, max(price) as max_price from painting group by a_id) as tmp    -> ON a.a_id = p.a_id    -> AND p.a_id = tmp.a_id    -> AND p.price = tmp.max_price;+----------+-------------------+-------+| name     | title             | price |+----------+-------------------+-------+| Da Vinci | The Mona Lisa     |    87 || Van Gogh | The Potato Eaters |    67 || Renoir   | Les Deux Soeurs   |    64 |+----------+-------------------+-------+

c——示例

mysql> select a.name, p1.title, p1.price    -> from painting as p1 LEFT JOIN painting as p2    -> ON p1.a_id = p2.a_id    -> AND p1.price < p2.price    -> INNER JOIN artist as a    -> ON p1.a_id = a.a_id    -> WHERE p2.a_id is null;+----------+-------------------+-------+| name     | title             | price |+----------+-------------------+-------+| Da Vinci | The Mona Lisa     |    87 || Renoir   | Les Deux Soeurs   |    64 || Van Gogh | The Potato Eaters |    67 |+----------+-------------------+-------+

6、计算小组积分榜

（Recipe 12.7. Computing Team Standings）

如下数据：

mysql> SELECT team, wins, losses FROM standings1    -> ORDER BY wins-losses DESC;+-------------+------+--------+| team        | wins | losses |+-------------+------+--------+| Winnipeg    |   37 |     20 || Crookston   |   31 |     25 || Fargo       |   30 |     26 || Grand Forks |   28 |     26 || Devils Lake |   19 |     31 || Cavalier    |   15 |     32 |+-------------+------+--------+

计算获胜的比例公式： wins / (wins + losses)

为了预防还没有一场比赛的情况，公式增强改为：IF( wins=0, 0, wins/(wins+losses) )

Game Behind 公式：( (winsA - lossesA) - (winsB - lossesB) ) / 2

给定一个小组（其实就是“Winnipeg”小组）作为比较的基准：

mysql> SET @wl_diff = (SELECT MAX(wins-losses) FROM standings1);

查询计算结果为：

mysql> SELECT team, wins AS W, losses AS L,    -> wins/(wins+losses) AS PCT,    -> (@wl_diff - (wins-losses)) / 2 AS GB    -> FROM standings1    -> ORDER BY wins-losses DESC, PCT DESC;+-------------+------+------+--------+---------+| team        | W    | L    | PCT    | GB      |+-------------+------+------+--------+---------+| Winnipeg    |   37 |   20 | 0.6491 |  0.0000 || Crookston   |   31 |   25 | 0.5536 |  5.5000 || Fargo       |   30 |   26 | 0.5357 |  6.5000 || Grand Forks |   28 |   26 | 0.5185 |  7.5000 || Devils Lake |   19 |   31 | 0.3800 | 14.5000 || Cavalier    |   15 |   32 | 0.3191 | 17.0000 |+-------------+------+------+--------+---------+

改进：

mysql> SELECT team, wins AS W, losses AS L,    -> TRUNCATE(wins/(wins+losses),3) AS PCT,    -> IF(@wl_diff = wins-losses,    ->    '-',TRUNCATE((@wl_diff - (wins-losses))/2,1)) AS GB    -> FROM standings1    -> ORDER BY wins-losses DESC, PCT DESC;+-------------+------+------+-------+------+| team        | W    | L    | PCT   | GB   |+-------------+------+------+-------+------+| Winnipeg    |   37 |   20 | 0.649 | -    || Crookston   |   31 |   25 | 0.553 | 5.5  || Fargo       |   30 |   26 | 0.535 | 6.5  || Grand Forks |   28 |   26 | 0.518 | 7.5  || Devils Lake |   19 |   31 | 0.380 | 14.5 || Cavalier    |   15 |   32 | 0.319 | 17.0 |+-------------+------+------+-------+------+

7、计算连续行的差

（Recipe 12.9. Calculating Successive-Row Differences）

mysql> SELECT seq, city, miles FROM trip_log ORDER BY seq;+-----+------------------+-------+| seq | city             | miles |+-----+------------------+-------+|   1 | San Antonio, TX  |     0 ||   2 | Dallas, TX       |   263 ||   3 | Benton, AR       |   566 ||   4 | Memphis, TN      |   745 ||   5 | Portageville, MO |   878 ||   6 | Champaign, IL    |  1164 ||   7 | Madison, WI      |  1412 |+-----+------------------+-------+

计算连续两城市的距离（dist）：

mysql> SELECT t1.seq AS seq1, t2.seq AS seq2,    -> t1.city AS city1, t2.city AS city2,    -> t1.miles AS miles1, t2.miles AS miles2,    -> t2.miles-t1.miles AS dist    -> FROM trip_log AS t1 INNER JOIN trip_log AS t2    -> ON t1.seq+1 = t2.seq    -> ORDER BY t1.seq;+------+------+------------------+------------------+--------+--------+------+| seq1 | seq2 | city1            | city2            | miles1 | miles2 | dist |+------+------+------------------+------------------+--------+--------+------+|    1 |    2 | San Antonio, TX  | Dallas, TX       |      0 |    263 |  263 ||    2 |    3 | Dallas, TX       | Benton, AR       |    263 |    566 |  303 ||    3 |    4 | Benton, AR       | Memphis, TN      |    566 |    745 |  179 ||    4 |    5 | Memphis, TN      | Portageville, MO |    745 |    878 |  133 ||    5 |    6 | Portageville, MO | Champaign, IL    |    878 |   1164 |  286 ||    6 |    7 | Champaign, IL    | Madison, WI      |   1164 |   1412 |  248 |+------+------+------------------+------------------+--------+--------+------+

8、计算“累计和”与运行时平均值

（Recipe 12.10. Finding Cumulative Sums and Running Averages）

mysql> SELECT date, precip FROM rainfall ORDER BY date;+------------+--------+| date       | precip |+------------+--------+| 2006-06-01 |   1.50 || 2006-06-02 |   0.00 || 2006-06-03 |   0.50 || 2006-06-04 |   0.00 || 2006-06-05 |   1.00 |+------------+--------+

计算每天的累计降雨量（自连接）：

mysql> SELECT t1.date, t1.precip AS 'daily precip',    -> SUM(t2.precip) AS 'cum. precip'    -> FROM rainfall AS t1 INNER JOIN rainfall AS t2    -> ON t1.date >= t2.date    -> GROUP BY t1.date;+------------+--------------+-------------+| date       | daily precip | cum. precip |+------------+--------------+-------------+| 2006-06-01 |         1.50 |        1.50 || 2006-06-02 |         0.00 |        1.50 || 2006-06-03 |         0.50 |        2.00 || 2006-06-04 |         0.00 |        2.00 || 2006-06-05 |         1.00 |        3.00 |+------------+--------------+-------------+

一般化：

mysql> SELECT t1.date, t1.precip AS 'daily precip',    -> SUM(t2.precip) AS 'cum. precip',    -> COUNT(t2.precip) AS 'days elapsed',    -> AVG(t2.precip) AS 'avg. precip'    -> FROM rainfall AS t1 INNER JOIN rainfall AS t2    -> ON t1.date >= t2.date    -> GROUP BY t1.date;+------------+--------------+-------------+--------------+-------------+| date       | daily precip | cum. precip | days elapsed | avg. precip |+------------+--------------+-------------+--------------+-------------+| 2006-06-01 |         1.50 |        1.50 |            1 |    1.500000 || 2006-06-02 |         0.00 |        1.50 |            2 |    0.750000 || 2006-06-03 |         0.50 |        2.00 |            3 |    0.666667 || 2006-06-04 |         0.00 |        2.00 |            4 |    0.500000 || 2006-06-05 |         1.00 |        3.00 |            5 |    0.600000 |+------------+--------------+-------------+--------------+-------------+

如果日期不连续呢（如，删除“precip”的值为 0 ，的行）

mysql> DELETE FROM rainfall WHERE precip = 0;mysql> SELECT date, precip FROM rainfall ORDER BY date;+------------+--------+| date       | precip |+------------+--------+| 2006-06-01 |   1.50 || 2006-06-03 |   0.50 || 2006-06-05 |   1.00 |+------------+--------+

计算累计天数 = DATEDIFF( MAX(t2.date), MIN(t2.date) ) + 1

计算平均降雨量 = 总的降雨量 / 累计天数

mysql> SELECT t1.date, t1.precip AS 'daily precip',    -> SUM(t2.precip) AS 'cum. precip',    -> DATEDIFF(MAX(t2.date),MIN(t2.date)) + 1 AS 'days elapsed',    -> SUM(t2.precip) / (DATEDIFF(MAX(t2.date),MIN(t2.date)) + 1)    -> AS 'avg. precip'    -> FROM rainfall AS t1 INNER JOIN rainfall AS t2    -> ON t1.date >= t2.date    -> GROUP BY t1.date;+------------+--------------+-------------+--------------+-------------+| date       | daily precip | cum. precip | days elapsed | avg. precip |+------------+--------------+-------------+--------------+-------------+| 2006-06-01 |         1.50 |        1.50 |            1 |    1.500000 || 2006-06-03 |         0.50 |        2.00 |            3 |    0.666667 || 2006-06-05 |         1.00 |        3.00 |            5 |    0.600000 |+------------+--------------+-------------+--------------+-------------+

另一例（对时间的特殊处理）

mysql> SELECT stage, km, t FROM marathon ORDER BY stage;+-------+----+----------+| stage | km | t        |+-------+----+----------+|     1 |  5 | 00:15:00 ||     2 |  7 | 00:19:30 ||     3 |  9 | 00:29:20 ||     4 |  5 | 00:17:50 |+-------+----+----------+

mysql> SELECT t1.stage, t1.km, t1.t,    -> SUM(t2.km) AS 'cum. km',    -> SEC_TO_TIME(SUM(TIME_TO_SEC(t2.t))) AS 'cum. t',    -> SUM(t2.km)/(SUM(TIME_TO_SEC(t2.t))/(60*60)) AS 'avg. km/hour'    -> FROM marathon AS t1 INNER JOIN marathon AS t2    -> ON t1.stage >= t2.stage    -> GROUP BY t1.stage;+-------+----+----------+---------+----------+--------------+| stage | km | t        | cum. km | cum. t   | avg. km/hour |+-------+----+----------+---------+----------+--------------+|     1 |  5 | 00:15:00 |       5 | 00:15:00 |      20.0000 ||     2 |  7 | 00:19:30 |      12 | 00:34:30 |      20.8696 ||     3 |  9 | 00:29:20 |      21 | 01:03:50 |      19.7389 ||     4 |  5 | 00:17:50 |      26 | 01:21:40 |      19.1020 |+-------+----+----------+---------+----------+--------------+

9、使用 JOIN 控制查询结果的顺序

（Recipe 12.11. Using a Join to Control Query Output Order）

a、创建一个辅助表，然后连接查询

b、使用 FROM 子句连接查询

数据如下：

mysql> SELECT * FROM driver_log ORDER BY rec_id;+--------+-------+------------+-------+| rec_id | name  | trav_date  | miles |+--------+-------+------------+-------+|      1 | Ben   | 2006-08-30 |   152 ||      2 | Suzi  | 2006-08-29 |   391 ||      3 | Henry | 2006-08-29 |   300 ||      4 | Henry | 2006-08-27 |    96 ||      5 | Ben   | 2006-08-29 |   131 ||      6 | Henry | 2006-08-26 |   115 ||      7 | Suzi  | 2006-09-02 |   502 ||      8 | Henry | 2006-09-01 |   197 ||      9 | Ben   | 2006-09-02 |    79 ||     10 | Henry | 2006-08-30 |   203 |+--------+-------+------------+-------+

任务：统计出每个人的总里程，并按总里程从大到小排序

a——示例

mysql> CREATE TABLE tmp    -> SELECT name, SUM(miles) AS driver_miles FROM driver_log GROUP BY name;    //演示——STARTmysql> SELECT * FROM tmp ORDER BY driver_miles DESC;+-------+--------------+| name  | driver_miles |+-------+--------------+| Henry |          911 || Suzi  |          893 || Ben   |          362 |+-------+--------------+//演示——ENDmysql> SELECT tmp.driver_miles, driver_log.*    -> FROM driver_log INNER JOIN tmp    -> ON driver_log.name = tmp.name    -> ORDER BY tmp.driver_miles DESC, driver_log.trav_date;+--------------+--------+-------+------------+-------+| driver_miles | rec_id | name  | trav_date  | miles |+--------------+--------+-------+------------+-------+|          911 |      6 | Henry | 2006-08-26 |   115 ||          911 |      4 | Henry | 2006-08-27 |    96 ||          911 |      3 | Henry | 2006-08-29 |   300 ||          911 |     10 | Henry | 2006-08-30 |   203 ||          911 |      8 | Henry | 2006-09-01 |   197 ||          893 |      2 | Suzi  | 2006-08-29 |   391 ||          893 |      7 | Suzi  | 2006-09-02 |   502 ||          362 |      5 | Ben   | 2006-08-29 |   131 ||          362 |      1 | Ben   | 2006-08-30 |   152 ||          362 |      9 | Ben   | 2006-09-02 |    79 |+--------------+--------+-------+------------+-------+

b——示例

mysql> SELECT tmp.driver_miles, driver_log.*    -> FROM driver_log INNER JOIN    -> (SELECT name, SUM(miles) AS driver_miles FROM driver_log GROUP BY name) AS tmp    -> ON driver_log.name = tmp.name    -> ORDER BY tmp.driver_miles DESC, driver_log.trav_date;+--------------+--------+-------+------------+-------+| driver_miles | rec_id | name  | trav_date  | miles |+--------------+--------+-------+------------+-------+|          911 |      6 | Henry | 2006-08-26 |   115 ||          911 |      4 | Henry | 2006-08-27 |    96 ||          911 |      3 | Henry | 2006-08-29 |   300 ||          911 |     10 | Henry | 2006-08-30 |   203 ||          911 |      8 | Henry | 2006-09-01 |   197 ||          893 |      2 | Suzi  | 2006-08-29 |   391 ||          893 |      7 | Suzi  | 2006-09-02 |   502 ||          362 |      5 | Ben   | 2006-08-29 |   131 ||          362 |      1 | Ben   | 2006-08-30 |   152 ||          362 |      9 | Ben   | 2006-09-02 |    79 |+--------------+--------+-------+------------+-------+

10、通过 UNION 合并查询结果集

（Recipe 12.12. Combining Several Result Sets in a Single Query）

UNION，删除重复的行

UNION ALL，不删除重复的行

演示数据：

mysql> SELECT * FROM prospect;+---------+-------+------------------------+| fname   | lname | addr                   |+---------+-------+------------------------+| Peter   | Jones | 482 Rush St., Apt. 402 || Bernice | Smith | 916 Maple Dr.          |+---------+-------+------------------------+mysql> SELECT * FROM customer;+-----------+------------+---------------------+| last_name | first_name | address             |+-----------+------------+---------------------+| Peterson  | Grace      | 16055 Seminole Ave. || Smith     | Bernice    | 916 Maple Dr.       || Brown     | Walter     | 8602 1st St.        |+-----------+------------+---------------------+mysql> SELECT * FROM vendor;+-------------------+---------------------+| company           | street              |+-------------------+---------------------+| ReddyParts, Inc.  | 38 Industrial Blvd. || Parts-to-go, Ltd. | 213B Commerce Park. |+-------------------+---------------------+

UNION——演示

mysql> SELECT fname, lname, addr FROM prospect    -> UNION    -> SELECT first_name, last_name, address FROM customer    -> UNION    -> SELECT company, '', street FROM vendor;+-------------------+----------+------------------------+| fname             | lname    | addr                   |+-------------------+----------+------------------------+| Peter             | Jones    | 482 Rush St., Apt. 402 || Bernice           | Smith    | 916 Maple Dr.          || Grace             | Peterson | 16055 Seminole Ave.    || Walter            | Brown    | 8602 1st St.           || ReddyParts, Inc.  |          | 38 Industrial Blvd.    || Parts-to-go, Ltd. |          | 213B Commerce Park.    |+-------------------+----------+------------------------+

UNION ALL——演示

mysql> SELECT fname, lname, addr FROM prospect    -> UNION ALL    -> SELECT first_name, last_name, address FROM customer    -> UNION ALL    -> SELECT company, '', street FROM vendor;+-------------------+----------+------------------------+| fname             | lname    | addr                   |+-------------------+----------+------------------------+| Peter             | Jones    | 482 Rush St., Apt. 402 || Bernice           | Smith    | 916 Maple Dr.          || Grace             | Peterson | 16055 Seminole Ave.    || Bernice           | Smith    | 916 Maple Dr.          || Walter            | Brown    | 8602 1st St.           || ReddyParts, Inc.  |          | 38 Industrial Blvd.    || Parts-to-go, Ltd. |          | 213B Commerce Park.    |+-------------------+----------+------------------------+

列的合并处理：

mysql> (SELECT CONCAT(lname,', ',fname) AS name, addr FROM prospect)    -> UNION    -> (SELECT CONCAT(last_name,', ',first_name), address FROM customer)    -> UNION    -> (SELECT company, street FROM vendor)    -> ORDER BY name;+-------------------+------------------------+| name              | addr                   |+-------------------+------------------------+| Brown, Walter     | 8602 1st St.           || Jones, Peter      | 482 Rush St., Apt. 402 || Parts-to-go, Ltd. | 213B Commerce Park.    || Peterson, Grace   | 16055 Seminole Ave.    || ReddyParts, Inc.  | 38 Industrial Blvd.    || Smith, Bernice    | 916 Maple Dr.          |+-------------------+------------------------+