假设我有桌子学生和桌子导师。学生包含ID和姓名:
+----+------+
| ID | name |
+----+------+advisor包含s_id和i_id:
+------+------+
| s_id | i_id |
+------+------+结果如何连接,如下所示:
+----+------+------+
| ID | name | i_id |
+----+------+------+我试过了
select *
from student join advisor on student.ID=advisor.s_id但它会返回
+----+------+------+------+
| ID | name | i_id | s_id |
+----+------+------+------+谢谢!
发布于 2020-11-09 10:13:08
修改现有查询,方法是在选择列表中仅包含您实际想要查看的列:
SELECT s.ID, s.name, a.i_id
FROM student s
INNER JOIN advisor a
on s.ID = a.s_id;如果给定的学生实际上没有导师,您可以在此处使用左连接来报告该学生的NA:
SELECT s.ID, s.name, COALESCE(a.i_id, 'NA') AS i_id
FROM student s
LEFT JOIN advisor a
on s.ID = a.s_id;https://stackoverflow.com/questions/64744813
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