Question

有人可以告诉我，我在哪里出错：

我在执行此操作时遇到错误：

CREATE PROCEDURE dbo.EnrollStudent ( @CourseID  AS INTEGER,
                                     @StudentID AS VARCHAR(20), 
                                     @Status AS VARCHAR(50) OUTPUT ) AS
BEGIN
   DECLARE @StutusID INTEGER

   IF @StutusID = 1
   BEGIN  
       SET @Status = 'The Student is already Enrolled'
   END;

   ELSE IF @StutusID = 2
   BEGIN 
       SET @Status = 'Cannot enroll until faculty is selected' 
   END

   ELSE IF @StutusID = 3
   BEGIN 
       SET @Status = 'Student Enrolled' 
   END

   IF (dbo.CourseEnrollment.CourseId = @CourseID AND 
       dbo.CourseEnrollment.StudentId LIKE @StudentID)
   BEGIN
      SET @StutusID = 1
   END
END;

我的错误是

Msg 4104，Level 16，State 1，Procedure EnrollStudent，Line 23
  多部分标识符＆＃34; dbo.CourseEnrollment.CourseId＆＃34;无法受约束   Msg 4104，Level 16，State 1，Procedure EnrollStudent，Line 23
  多部分标识符＆＃34; dbo.CourseEnrollment.StudentId＆＃34;无法绑定

如何从CourseEnrollment表格中进行选择？我无法做到这一点。

Answer 1

您不能在不使用表格的情况下使用表格列：

IF EXISTS (SELECT 1 FROM dbo.CourseEnrollment WHERE CourseId=@CourseID 
           AND StudentId LIKE '%' + @StudentID + '%')
BEGIN
SET @StutusID=1
END

Answer 2

Yout Code有点不清楚。在声明并且未设置变量后，检查@stutusID是否为1。

此后，您可以阅读表格的一些数据。

我认为您正在尝试检查表格中是否已有学生。你应该做那样的事情：

CREATE PROCEDURE dbo.EnrollStudent ( @CourseID  AS INTEGER,
                                 @StudentID AS VARCHAR(20), 
                                 @Status AS VARCHAR(50) OUTPUT ) AS
BEGIN
   DECLARE @StutusID INTEGER

   SELECT COUNT(*) INTO @StutisID FROM dbo.CourseEnrollment WHERE dbo.CourseEnrollment.CourseId = @CourseID AND 
       dbo.CourseEnrollment.StudentId LIKE @StudentID

   IF @StutusID = 1
   BEGIN  
       SET @Status = 'The Student is already Enrolled'
   END;

       ELSE IF @StutusID = 2
   BEGIN 
       SET @Status = 'Cannot enroll until faculty is selected' 
   END

   ELSE IF @StutusID = 3
   BEGIN 
       SET @Status = 'Student Enrolled' 
   END
END;

存储过程无法选择表格

2 个答案: