如何将多个表中的一条记录组合起来

Question

我有两个表课程和科目。这是一个一对多的比率在每节课可以在导致类似“名称”主题1个或多个科目进行分类：数学，科学，社会研究

这将返回3个记录：
名称，数学
名称，科学
名，社会科学

这导致我通过循环和组合以及打乱了我的搜索计数
我怎样才能得到一个记录所有三个呢？所以我的搜索计数是准确的，我不需要额外的PHP代码来检查名称是否相同并收集额外的主题？

我试过基本

SELECT * FROM lesson INNER JOIN subjects ON subject.id = lesson.subject 

但是这导致3项

编辑
我的查询更复杂，比我带领。我有一个中间表，用于跟踪上面的两个表格及其关系
此查询用于搜索。这是我的。

SELECT name, subject   
FROM lesson As l 
INNER JOIN lesson_sub As ls 
    ON ls.lesson_id = l.id 
INNER JOIN subjects As s 
    ON s.id = ls.subject_id 
WHERE CONCAT(l.name, s.subject) LIKE '%KEYWORD%' AND s.id = SUBJECT_ID 

Answer 1

可以使用GROUP_CONCAT（）函数与JOIN查询：

http://dev.mysql.com/doc/refman/5.0/en/group-by-functions.html#function_group-concat

GROUP_CONCAT（expr）

该函数返回一个字符串结果，其中包含来自一个组的连接的非NULL 值。如果没有非NULL值，它将返回NULL。 完整的语法如下：

GROUP_CONCAT（[DISTINCT] expr的[，... EXPR] [ORDER BY {UNSIGNED_INTEGER | COL_NAME | expr}的 [ASC | DESC] [，... COL_NAME]] [SEPARATOR str_val]）

的MySQL> SELECT student_name， - > GROUP_CONCAT（test_score） - > FROM学生 - > GROUP BY student_name;

或者：

的mysql> SELECT student_name， - > GROUP_CONCAT（DISTINCT test_score - > ORDER BY test_score DESC SEPARATOR'“） - >从学生 - > GROUP BY student_name;

所以：

SELECT lesson.name, GROUP_CONCAT(subjects.name SEPARATOR ', ') 
    FROM lesson JOIN subjects ON (subject.id = lesson.subject) 
    GROUP BY lesson.name; 

TEST

CREATE TABLE lesson (name varchar (20), subject integer); 
CREATE TABLE subjects (id integer, name varchar(20)); 

INSERT INTO subjects VALUES (1, 'Math'), (2, 'Physics'), (3, 'Chemistry'); 

INSERT INTO lesson VALUES ('Lesson A', 1); 
INSERT INTO lesson VALUES ('Lesson A', 2); 
INSERT INTO lesson VALUES ('Lesson A', 3); 
INSERT INTO lesson VALUES ('Lesson B', 2); 
INSERT INTO lesson VALUES ('Lesson B', 3); 
INSERT INTO lesson VALUES ('Lesson C', 1); 

SELECT lesson.name, GROUP_CONCAT(subjects.name SEPARATOR ', ') 
    FROM lesson JOIN subjects ON (subjects.id = lesson.subject) 
    GROUP BY lesson.name; 

+----------+--------------------------------------------+ 
| name  | GROUP_CONCAT(subjects.name SEPARATOR ', ') | 
+----------+--------------------------------------------+ 
| Lesson A | Math, Chemistry, Physics     | 
| Lesson B | Chemistry, Physics       | 
| Lesson C | Math          | 
+----------+--------------------------------------------+ 
3 rows in set (0.00 sec) 

更复杂的TEST（具有中间表）

CREATE TABLE lesson (id integer, name varchar (20)); 
CREATE TABLE subjects (id integer, name varchar(20)); 
CREATE TABLE lesson_sub (lesson_id integer, subject_id integer); 

INSERT INTO subjects VALUES (1, 'Math'), (2, 'Physics'), (3, 'Chemistry'); 
INSERT INTO lesson VALUES (1, 'Lesson A'), (2, 'Lesson B'), (3, 'Lesson C'); 

INSERT INTO lesson_sub VALUES (1,1), (1,2),(1,3),(2,2),(2,3),(3,1); 

SELECT lesson.name, GROUP_CONCAT(subjects.name SEPARATOR ', ') AS subjects 
    FROM lesson_sub JOIN lesson ON (lesson.id = lesson_sub.lesson_id) 
        JOIN subjects ON (subjects.id = lesson_sub.subject_id) 
    WHERE CONCAT(lesson.name, subjects.name) LIKE '%Chem%' 
    GROUP BY lesson.name; 

SELECT name, subjects FROM (
    SELECT lesson.name, GROUP_CONCAT(subjects.name SEPARATOR ', ') AS subjects 
    FROM lesson_sub JOIN lesson ON (lesson.id = lesson_sub.lesson_id) 
        JOIN subjects ON (subjects.id = lesson_sub.subject_id) 
    GROUP BY lesson.name) AS lesson_clear 
    WHERE CONCAT(name, subjects) LIKE '%Chem%'; 

+----------+--------------------------------------------+ 
| name  | GROUP_CONCAT(subjects.name SEPARATOR ', ') | 
+----------+--------------------------------------------+ 
| Lesson A | Chemistry         | 
| Lesson B | Chemistry         | 
+----------+--------------------------------------------+ 
2 rows in set (0.00 sec) 

+----------+--------------------------+ 
| name  | subjects     | 
+----------+--------------------------+ 
| Lesson A | Physics, Math, Chemistry | 
| Lesson B | Physics, Chemistry  | 
+----------+--------------------------+ 
2 rows in set (0.00 sec) 

Answer 2

为什么内连接，如果你只是想要主题名称？现在

SELECT subject 
FROM lesson 

，如果你想要的是至少有一个课科目，那么你可以简单地做

SELECT DISTINCT subject 
FROM lesson 
INNER JOIN subjects ON subject.id = lesson.subject 

Answer 3

它升像你这样的人希望每人有一个记录，那就是每name。

这里有一个办法做到这一点：

 SELECT name, count(*) subject_count 
     FROM lesson 
    INNER JOIN subjects ON subject.id = lesson.subject 
    GROUP BY name 
    ORDER BY name 

如果需要在名字的记录中显示的主题，试试这个：

 SELECT name, 
      count(*) subject_count, 
      group_concat(subject) subjects 
     FROM lesson 
    INNER JOIN subjects ON subject.id = lesson.subject 
    GROUP BY name 
    ORDER BY name