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

———————————————————————————————————————————

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 IDname  VARCHAR(30) NOT NULL,                 # artist namePRIMARY KEY (a_id),UNIQUE (name)
);CREATE TABLE painting
(a_id  INT UNSIGNED NOT NULL,                # artist IDp_id  INT UNSIGNED NOT NULL AUTO_INCREMENT, # painting IDtitle VARCHAR(100) NOT NULL,                # title of paintingstate VARCHAR(2) NOT NULL,                  # state where purchasedprice 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-START
mysql> 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——START
mysql> 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.          |
+-------------------+------------------------+