Question

当我运行以下SIMPLE代码时，我收到以下错误：＆＃34;必须声明表变量＆＃34; @ tableName＆＃34;。＆＃34;

我不确定原因：我在WHILE语句中尝试DECLAREing变量，但我仍然得到同样的愚蠢错误。我需要做的就是查询我的数据库（myDB）中的所有表并输出它。

DECLARE @tableName VARCHAR(80)
DECLARE TableName_Cursor CURSOR FAST_FORWARD FOR
    SELECT TABLE_NAME FROM myDB.INFORMATION_SCHEMA.Tables 

OPEN TableName_Cursor

FETCH NEXT FROM TableName_Cursor INTO @tableName

WHILE @@FETCH_STATUS =0
BEGIN
--DECLARE @tableName VARCHAR(80)
--USE myDB
    SELECT * FROM @tableName
    FETCH NEXT FROM TableName_Cursor INTO @tableName
END
CLOSE TableName_Cursor
DEALLOCATE TableName_Cursor

非常感谢你的帮助！

Answer 1

与variable范围无关

您需要Dynamic SQL才能执行此操作

Declare @sql varchar(8000)= ''
WHILE @@FETCH_STATUS =0
BEGIN
    Set @sql = ''     
    Set @sql = 'SELECT * FROM ' +QUOTENAME(@tableName)
    EXEC(@sql)
    FETCH NEXT FROM TableName_Cursor INTO @tableName
END

SQL Server中游标的可变范围？

1 个答案: