Question

    DECLARE sp_ApplicationId char(38);
    SET sp_ApplicationId = UUID();

   IF(EXISTS(SELECT sp_ApplicationId = ApplicationId FROM aspnet_Applications WHERE       LOWER(sp_ApplicationName) = LoweredApplicationName)) THEN

               SELECT 'Application Name Already Exist';

我想在满足条件时将应用程序ID分配给sp_ApplicationId我在这里使用=运算符，但它只比较值。我也试过：=运算符，但它显示以下错误：

 Error code 1064, SQL state 42000: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ':= ApplicationId FROM aspnet_Applications WHERE LOWER(sp_ApplicationName) = Lowe' at line 14

请帮我解决这个问题。

Answer 1

尝试更改：

SELECT sp_ApplicationId = ApplicationId FROM aspnet_Applications WHERE       LOWER(sp_ApplicationName) = LoweredApplicationName

为：

SELECT 1 FROM aspnet_Applications WHERE       LOWER(sp_ApplicationName) = LoweredApplicationName

更新如果您尝试将ApplicationId的值分配给sp_ApplicationId变量，可以尝试：

SELECT ApplicationId INTO sp_ApplicationId FROM aspnet_Applications WHERE       LOWER(sp_ApplicationName) = LoweredApplicationName

如何在mysql存储过程中使用赋值运算符

1 个答案: