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我有两张桌子课和科目。这是一个一对多的比例，因为在每节课中都可以归类为 1 个或多个科目，从而产生类似于“名称”主题的内容：数学、科学、社会研究

这将返回 3 条记录：
姓名、数学
名称、科学
名称、社会研究

这导致我循环并组合并弄乱了我的搜索计数
我怎样才能获得所有三个记录的一条记录？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？

我试过基本的

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

但这会导致 3 个条目

编辑
我的查询比我引导的要复杂。我有一个中间表跟踪上面的两个表及其关系
这个查询用于搜索。这就是我所拥有的。

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
4

3 回答 3

2

您可以在 JOIN 查询中使用 GROUP_CONCAT() 函数：

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

GROUP_CONCAT(expr)

此函数返回一个字符串结果，其中包含来自组的串联非 NULL 值。如果没有非 NULL 值，则返回 NULL。完整的语法如下：

GROUP_CONCAT([DISTINCT] expr [,expr ...] [ORDER BY {unsigned_integer | col_name | expr} [ASC | DESC] [,col_name ...]] [SEPARATOR str_val])

mysql> SELECT student_name, -> GROUP_CONCAT(test_score) -> FROM student -> GROUP BY student_name;

或者：

mysql> SELECT student_name, -> GROUP_CONCAT(DISTINCT test_score -> ORDER BY test_score DESC SEPARATOR ' ') -> FROM student -> 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;

测试

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)

更复杂的测试（带中间表）

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)
于 2012-08-22T16:06:58.960 回答
0

如果您只想要主题名称，为什么要使用内部连接？

SELECT subject
FROM lesson

现在，如果你想要至少有一节课的科目，那么你可以简单地做

SELECT DISTINCT subject
FROM lesson
INNER JOIN subjects ON subject.id = lesson.subject
于 2012-08-22T16:04:57.800 回答
0

看起来您希望每人有一个记录，即 per name。

这是一种方法：

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

如果您需要按名称显示在记录中的主题，请尝试以下操作：

      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
于 2012-08-22T16:11:15.390 回答