Question

Problem: The code reads each row and should determine whether it is empty or not. I want to print out "works" if the column is null. Otherwise, it should not do anything. Right now it is not working. It is giving me an error that the column does not exist.(the one mentioned in the if statement)

Here is the code. Code:

DECLARE Employee_Cursor CURSOR FOR

SELECT [Description]  
FROM tblSubViews ts  
INNER JOIN tblViewSubViewJoin tvs  
ON ts.SubViewId = tvs.SubViewId Where tvs.ViewId = 4

OPEN Employee_Cursor

FETCH NEXT FROM Employee_Cursor
WHILE @@FETCH_STATUS = 0

BEGIN   
if (Description = null)
begin
print  works 
end
   FETCH NEXT FROM Employee_Cursor
END
CLOSE Employee_Cursor
DEALLOCATE Employee_Cursor

Answer 1

您不能与平等使用的<代码>NUL进行比较。

在使用治疗器时,还必须选择成一个变量:

  FETCH NEXT FROM Employee_Cursor INTO @Description 
  WHILE @@FETCH_STATUS = 0

当然,你需要宣布@描述为适合该栏的类型,

Answer 2

。

因此,鉴于其价值并不高,你不能与经常手段进行比较——你需要使用<代码>。 ISNUL or IS NOT NUL methods:

if (Description IS NULL)
begin
  print  works 
end

与<代码>IS NUL或. IS NOT NUL相比,你将重新研究结果。

Answer 3

在这种情况下,迫切需要一名治疗员。

SELECT 
[Description],
ISNULL([Description]   , Works ) AS text
FROM tblSubViews ts   
INNER JOIN 
tblViewSubViewJoin tvs   
ON 
ts.SubViewId = tvs.SubViewId 
Where 
tvs.ViewId = 4

那么,你可以通过在你的《经济、社会、文化权利国际公约》条款中增加一项额外标准来缩短这一节点,而这一条款只选择了无说明的栏目?

Answer 4

根据你的逻辑,在你的询问中使用<代码>snull()。

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