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,它由上述步骤中连接的表名组成。此语句没有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?
最佳答案
默认情况下,您无法参数化表名和列名,因此您需要为此创建动态 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;
关于mysql - 如何在select语句中获取变量值作为表名,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/14828838/