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insert语句中有一个名为“Project_List_val”的变量，它由在上一步中合并的表名组成。此语句不将变量的内容作为表名，而是将“Project_List_val”作为表名并给出表未找到错误。

有什么建议吗？

Answer 1

默认您无法参数化表名和列名，因此您需要为其创建动态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;

如何在select语句中将变量值作为表名获取

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