VB报错index must be greater than zero and less than or equal t

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VB报错index must be greater than zero and less than or equal to count
sql = "Insert into HaiTian (加工号,工件号,程序号,开始日期,开始时间,状态)" & _
       " VALUES( '" & MachineName & "','" & Numbe & "','" & _
      ProgramName & "','" & Date & "','" & Time & _
       "','未完成')"

帮我看下,怎么一运行就报这个错!

你把sql打印出来看看是什么.初步怀疑是类型和DB定义的不一致导致的.
再问：额，看了下，确实是类型没对上。
再问下比如我有两个时间的字符串：
DateTime1 = "2011年6月27日 6:52:22"
DateTime2 = “2011年6月28日 13:52:14"
我怎么求出这两个时间的中间时长
再答：如果是sql的话，可以使用datediff函数
如果是oracle的话，可以使用下面的sql。
select floor(to_number(sysdate-to_date('2007-11-02 15:55:03','yyyy-mm-dd hh24:mi:ss'))/365) as spanYears from dual //??差-年
select ceil(moths_between(sysdate-to_date('2007-11-02 15:55:03','yyyy-mm-dd hh24:mi:ss'))) as spanMonths from dual //??差-月
select floor(to_number(sysdate-to_date('2007-11-02 15:55:03','yyyy-mm-dd hh24:mi:ss'))) as spanDays from dual //??差-天
select floor(to_number(sysdate-to_date('2007-11-02 15:55:03','yyyy-mm-dd hh24:mi:ss'))*24) as spanHours from dual //??差-?
select floor(to_number(sysdate-to_date('2007-11-02 15:55:03','yyyy-mm-dd hh24:mi:ss'))*24*60) as spanMinutes from dual //??差-分
select floor(to_number(sysdate-to_date('2007-11-02 15:55:03','yyyy-mm-dd hh24:mi:ss'))*24*60*60) as spanSeconds from dual //??差-秒

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