DECLARE @StartTime datetime ,
@s varchar(4000)SELECT @StartTime = '2004-4-13 20:00:00'SELECT @s='DATEDIFF(SECOND, ' + char(39) + cast(@StartTime as varchar(40)) + char(39) + ', 字段1) FROM 表1'select (@s)
@s varchar(4000)SELECT @StartTime = '2004-4-13 20:00:00'SELECT @s='DATEDIFF(SECOND, ' + char(39) + cast(@StartTime as varchar(40)) + char(39) + ', 字段1) FROM 表1'select (@s)
@s varchar(4000)SELECT @StartTime = '2004-4-13 20:00:00'SELECT @s='select --你的语句中为什么没有select ?
DATEDIFF(SECOND, ' + char(39) + cast(@StartTime as varchar(40)) + char(39) + ', 字段1) FROM 表1'print(@s) --查询分析器中显示@s的值
exec (@s)
SELECT @StartTime = '2004-4-13 20:00:00'
SELECT DATEDIFF(SECOND,@StartTime,字段1) FROM 表1
DECLARE @StartTime datetime ,
@s varchar(4000)
SELECT @StartTime = '2004-4-13 20:00:00'
SELECT @s='select DATEDIFF(SECOND,'''+@StartTime+''',字段1) FROM 表1'
exec (@s)
@s varchar(4000)SELECT @StartTime = '2004-4-13 20:00:00'SELECT @s='DATEDIFF(SECOND, ' + char(39) + cast(@StartTime as varchar(40)) + char(39) + ', 字段1)/ ' + @T + ' FROM 表1'
提示:
将 varchar 值 'DATEDIFF(SECOND, '04 13 2004 8:00PM', 字段1) /' 转换为数据类型为 int 的列时发生语法错误。请问何故?