Question

我有一个函数可以生成下面定义的评估信息类型。

create type course_assessment_info as (
  total_count           int,
  mark_average          double precision,
  weight_sum            double precision,
  weighted_mark_average double precision
);

当我将函数用于单个查询时，如下所示，我得到了预期的结果。

select * 
from course_assessment_info_by_student_id_and_course_id('9be15896-40ca-46c6-8fdd-0ffe3bd79
586', '65dbdce1-fd76-4951-9db1-d089b3794d80');

结果：

 total_count |   mark_average    |    weight_sum     | weighted_mark_average
-------------+-------------------+-------------------+-----------------------
           1 | 0.834768535328714 | 0.540032932289522 |     0.450802499916595

当我在使用联接的较大函数上使用它时，我没有得到单独的列名。我的理解是，我需要使用select * from course_asses ... 但是，我很难弄清楚如何格式化以使PostgreSQL不会引发语法错误。

select course_assessment_info_by_student_id_and_course_id(s.id, c.id)
from student s
join student_course sc on s.id = sc.student_id
join course c on c.id = sc.course_id;

感谢您的帮助！

Answer 1

您似乎想要LATERAL JOIN：

SELECT t.*
FROM student s
INNER JOIN student_course sc ON s.id = sc.student_id
INNER JOIN course c ON c.id = sc.course_id
INNER JOIN LATERAL course_assessment_info_by_student_id_and_course_id(s.id, c.id) t ON true

这也应该做：

select (course_assessment_info_by_student_id_and_course_id(s.id, c.id)).*
from student s
join student_course sc on s.id = sc.student_id
join course c on c.id = sc.course_id;

Sql Table从Postgres函数的结果中选择

1 个答案: