Question

id  Subject   mark   Year
-------------------------
1   Maths      32    2008
1   Science    40    2009
1   Science    45    2008
1   English    50    2009
1   English    60    2008

我正在寻找这样的结果：

id  Maths   Science   English
-----------------------------
1   32      40 & 45   50 & 60

请指教。我正在使用MySQL。

Answer 1

作为@Mark suggested，GROUP_CONCAT()可以为您提供以下结果：

SELECT   id, subject, GROUP_CONCAT(mark SEPARATOR ' & ') marks
FROM     results 
GROUP BY id, subject;
+------+---------+---------+
| id   | subject | marks   |
+------+---------+---------+
|    1 | English | 50 & 60 |
|    1 | Maths   | 32      |
|    1 | Science | 40 & 45 |
+------+---------+---------+
3 rows in set (0.00 sec)

从这个测试案例中：

CREATE TABLE results (id int, subject varchar(10), mark int);

INSERT INTO results VALUES (1, 'Maths', 32);
INSERT INTO results VALUES (1, 'Science', 40);
INSERT INTO results VALUES (1, 'Science', 45);
INSERT INTO results VALUES (1, 'English', 50);
INSERT INTO results VALUES (1, 'English', 60);

然而，另一种解决问题的方法是使用每个主题的子查询：

SELECT    r.id,
          (SELECT   GROUP_CONCAT(r_eng.mark SEPARATOR ' & ')
           FROM     results r_eng
           WHERE    r_eng.subject = 'English' AND r_eng.id = r.id) English,
          (SELECT   GROUP_CONCAT(r_eng.mark SEPARATOR ' & ') 
           FROM     results r_eng
           WHERE    r_eng.subject = 'Maths' AND r_eng.id = r.id) Maths,
          (SELECT   GROUP_CONCAT(r_eng.mark SEPARATOR ' & ') 
           FROM     results r_eng
           WHERE    r_eng.subject = 'Science' AND r_eng.id = r.id) Science
FROM      results r
GROUP BY  r.id;

将产生以下结果：

+------+---------+-------+---------+
| id   | English | Maths | Science |
+------+---------+-------+---------+
|    1 | 50 & 60 | 32    | 40 & 45 |
+------+---------+-------+---------+
1 row in set (0.01 sec)

<强>更新

除了评论之外，您还需要考虑year字段。幸运的是，GROUP_CONCAT()函数采用了我们可以使用的ORDER BY子句。让我们从一个带有年份字段的新测试用例开始：

CREATE TABLE results (id int, subject varchar(10), mark int, year int);

INSERT INTO results VALUES (1, 'Maths', 32, 2008);
INSERT INTO results VALUES (1, 'Science', 40, 2009);
INSERT INTO results VALUES (1, 'Science', 45, 2008);
INSERT INTO results VALUES (1, 'English', 50, 2009);
INSERT INTO results VALUES (1, 'English', 60, 2008);

SELECT * FROM results;
+------+---------+------+------+
| id   | subject | mark | year |
+------+---------+------+------+
|    1 | Maths   |   32 | 2008 |
|    1 | Science |   40 | 2009 |
|    1 | Science |   45 | 2008 |
|    1 | English |   50 | 2009 |
|    1 | English |   60 | 2008 |
+------+---------+------+------+
5 rows in set (0.00 sec)

然后我们可以将GROUP_CONCAT()函数与ORDER BY子句一起使用，如下所示：

SELECT   id, 
         subject, 
         GROUP_CONCAT(mark ORDER BY year SEPARATOR ' & ') marks, 
         GROUP_CONCAT(year ORDER BY year SEPARATOR ' & ') years
FROM     results 
GROUP BY id, subject;

+------+---------+---------+-------------+
| id   | subject | marks   | years       |
+------+---------+---------+-------------+
|    1 | English | 60 & 50 | 2008 & 2009 |
|    1 | Maths   | 32      | 2008        |
|    1 | Science | 45 & 40 | 2008 & 2009 |
+------+---------+---------+-------------+
3 rows in set (0.00 sec)

最后，对于GROUP BY一个水平行中的所有内容，我们可以使用前面示例中使用的子查询技术：

SELECT    r.id,
          (SELECT  GROUP_CONCAT(r_eng.mark ORDER BY year SEPARATOR ' & ') 
           FROM    results r_eng
           WHERE   r_eng.subject = 'English' AND r_eng.id = r.id) English,
          (SELECT  GROUP_CONCAT(r_eng.year ORDER BY year SEPARATOR ' & ')
            FROM   results r_eng
            WHERE  r_eng.subject = 'English' AND r_eng.id = r.id) Years_English,
          (SELECT  GROUP_CONCAT(r_eng.mark ORDER BY year SEPARATOR ' & ')
           FROM    results r_eng
           WHERE   r_eng.subject = 'Maths' AND r_eng.id = r.id) Maths,
          (SELECT  GROUP_CONCAT(r_eng.year ORDER BY year SEPARATOR ' & ')
            FROM   results r_eng
            WHERE  r_eng.subject = 'Maths' AND r_eng.id = r.id) Years_Maths,
          (SELECT  GROUP_CONCAT(r_eng.mark ORDER BY year SEPARATOR ' & ')
           FROM    results r_eng
           WHERE   r_eng.subject = 'Science' AND r_eng.id = r.id) Science,
          (SELECT  GROUP_CONCAT(r_eng.year ORDER BY year SEPARATOR ' & ')
            FROM   results r_eng
            WHERE  r_eng.subject = 'Science' AND r_eng.id = r.id) Years_Science
FROM      results r
GROUP BY  r.id;

将返回以下内容：

+----+---------+---------------+-------+-------------+---------+---------------+
| id | English | Years_English | Maths | Years_Maths | Science | Years_Science |
+----+---------+---------------+-------+-------------+---------+---------------+
|  1 | 60 & 50 | 2008 & 2009   | 32    | 2008        | 45 & 40 | 2008 & 2009   |
+----+---------+---------------+-------+-------------+---------+---------------+
1 row in set (0.01 sec)

如果您希望按降序排列标记和年份，则只需在每个DESC后添加ORDER BY year关键字。

Answer 2

第一步，看看mysql中的group_concat()功能，可能会有所帮助

SELECT `id`,
       `subject`,
       group_concat( `mark`
                     ORDER BY `year` DESC
                     SEPARATOR ' & '
                   ) marks
  FROM `subjects` 
 GROUP BY `id`,
          `subject`

<强>更新管理到目前为止

SELECT `id`,
       IF (`subject` = 'English', `marks`, NULL) AS English,
       IF (`subject` = 'Maths', `marks`, NULL) AS Maths,
       IF (`subject` = 'Science', `marks`, NULL) AS Science
  FROM ( SELECT `id`, 
                `subject`, 
                group_concat( `mark` 
                              ORDER BY `year` DESC 
                              SEPARATOR ' & ' 
                            ) marks 
           FROM `subjects`  
          GROUP BY `id`, 
                   `subject` 
         ) x
GROUP BY `id`,
         `subject`

返回

+----+---------+-------+---------+
| id | English | Maths | Science |
+----+---------+-------+---------+
|  1 | 50 & 60 | NULL  | NULL    |
|  1 | NULL    | 32    | NULL    |
|  1 | NULL    | NULL  | 40 & 45 |
+----+---------+-------+---------+
3 rows in set (0.01 sec)

但是不能完全获得最后一个分组权限将它全部作为一行返回。在采用这种方法解决问题时，不知道是否有其他人可以帮助我做到最后一步....通常，我会将结果返回给PHP“按原样”，并在那里进行最后的分组;但是一个完整的MySQL解决方案会很有趣。

Mysql：如何将多行连接成具有不同字段名称的单行？

2 个答案: