mysql - Mysql如何从选择结果中创建列名

Question

我想根据列中的子环值从不同的数据库中选择结果。这是我的表学生：

Original_student        Other_student
1010173                   1240240
1010173                   1240249

编号中的第 3 位将用于区分数据库。例如。我希望查询是

select original_student, Other_student, month
from student join database-(substring(other_student,3,1).payment 

我的问题是：如何将子字符串动态连接到数据库名称或列名称？

谢谢

Answer 1

假设您有一个字段来通过唯一的 id ( id_student) 来识别每个学生，这里有一个便宜的替代方案：

CREATE OR REPLACE VIEW v_student_payment AS
SELECT 0 AS db, payment, id_student FROM database-0
UNION 
SELECT 1 AS db, payment, id_student FROM database-1
UNION 
SELECT 2 AS db, payment, id_student FROM database-2
UNION 
SELECT 3 AS db, payment, id_student FROM database-3
/* here you have to add all databases you're using. There's a little maintenance cost, for if one day there's a new database to be created this view would have to be modified */
;

SELECT 
    original_student, 
    Other_student, 
    month, 
    v.payment
FROM 
    student s
        JOIN v_student_payment v ON v.id_student = s.id_student AND v.db = SUBSTRING(other_student,3,1) 

Answer 2

您是否尝试使用 Case 语句来检查连接。尝试查看此链接

Join 中的用例语句