有可能吗? 我想要得到的是这样的:
+-------------------------------+
| data 1 (another1, another2); |
| data 2 (another1 ..... ) |
+-------------------------------+
从表中可以看到这样的内容:
--------------------------------+
data1 | another1 | foreign key1 |
--------------------------------+
data1 | another2 | foreign key1 |
--------------------------------+
data2 | another1 | foreign key1 |
--------------------------------+
我使用这个代码
GROUP_CONCAT(DISTINCT CONCAT(data, ' (', another, ')') SEPARATOR ';\r\n ') as InOneCell
并得到这个输出:
+-------------------------------+
| data 1 (another1); |
| data 1 (another2); |
| data 2 (another1); ... |
+-------------------------------+
尝试过这个:
GROUP_CONCAT(DISTINCT CONCAT(data, ' (', GROUP_CONCAT(another SEPARATOR ', '), ')') SEPARATOR ';\r\n ') as InOneCell
并得到错误“组功能无效使用”,这真的很阴暗-_- 由于复杂的查询,我按 - 外键进行分组(下)。
我做错了什么或者仅在 SQL 上不可能?
这里是完整的查询:
SELECT
s.name,
s.surname,
gr.number AS 'group',
s.bitrh_date,
s.email,
s.registration_ip,
s.registration_time,
GROUP_CONCAT(DISTINCT CONCAT(dis.code, ' ', dis.title, ' (', GROUP_CONCAT(DISTINCT g.mark SEPARATOR ', '), ')') SEPARATOR ';\r\n ') as learning,
ROUND(AVG(g.mark), 2) AS average,
gr.semester,
ref.text AS reference
FROM t_students s
LEFT JOIN t_grades g
ON s.id = g.student
LEFT JOIN t_groups gr
ON s.group_id = gr.id
LEFT JOIN t_references ref
ON ref.author = s.id
LEFT JOIN t_program prog
ON gr.id = prog.group_id
LEFT JOIN t_disciplines dis
ON prog.discipline_id = dis.code
GROUP BY s.id
ORDER BY s.registration_time DESC
最佳答案
您应该使用group by(并且单个group_concat就足够了)
select CONCAT(data, ' (', GROUP_CONCAT(another SEPARATOR ', '), ');') as InOneCell
from my_table
group by data
或者您可以使用子查询
select data, group_concat(my_group)
from (
select data, group_concat(another) my_group
group by data
) t
group by data
就你的情况而言 您可以使用子查询,例如:
SELECT
s.name,
s.surname,
gr.number AS 'group',
s.bitrh_date,
s.email,
s.registration_ip,
s.registration_time,
GROUP_CONCAT(DISTINCT CONCAT(dis.code, ' ', dis.title, ' (', t.learning, ')') SEPARATOR ';\r\n ') as learning,
t.average,
gr.semester,
ref.text AS reference
FROM t_students s
LEFT JOIN (SELECT
student,
GROUP_CONCAT(DISTINCT mark SEPARATOR ', ') learning,
ROUND(AVG(mark), 2) AS average
from t_grades
group by student
) t on t.student = s.id
LEFT JOIN t_groups gr
ON s.group_id = gr.id
LEFT JOIN t_references ref
ON ref.author = s.id
LEFT JOIN t_program prog
ON gr.id = prog.group_id
LEFT JOIN t_disciplines dis
ON prog.discipline_id = dis.code
GROUP BY s.id
ORDER BY s.registration_time DESC
关于MySQL GROUP_CONCAT 位于另一个 GROUP_CONCAT 中,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/51234627/