mysql - 如何在select语句中获取变量值作为表名

Question

SET Project_List_val=CONCAT(Project_Number_val,'_List');
Insert Into test (Manthan_Panel_Id) select Manthan_Panel_Id from Project_List_val where Project_Number_val='9';

在插入语句中，有一个名为“ Project_List_val”的变量，该变量由上一步中确认的表名组成。该语句不将变量的内容作为表名，而是将“ Project_List_val”作为表名，并给出未找到表的错误。

有什么建议么？

Answer 1

默认情况下，您无法参数化表名和列名，因此需要为此创建Dynamic SQL，

SET @Project_List_val = CONCAT(Project_Number_val, '_List');
SET @projNum = 9;
SET @sql = CONCAT(' INSERT INTO test (Manthan_Panel_Id)
                    SELECT Manthan_Panel_Id 
                    FROM ', @Project_List_val, ' 
                    WHERE Project_Number_val = ?');

PREPARE stmt FROM @sql;
EXECUTE stmt USING @projNum;
DEALLOCATE PREPARE stmt;