Question

VB 怎么将list1.list(i)路径文件打开并赋值给a（i）?

For i = 0 To List2.ListCount - 1
open list2.list(i) For Binary Access Write As #2 'list2.list(0)=c:\1.txt
get #2, ,a(i)
close #2
Next

Answer 1

修改下楼上的代码：
Dim arr() As Byte, a()
ReDim a(list2.ListCount - 1)
For i = 0 To list2.ListCount - 1
Open list2.List(i) For Binary Access Read As #2 'list2.list(0)=c:\1.txt 把Write改成Read
Get #2, , arr
Close #2
a(i) = arr
Next

Answer 2

For i = 0 To List2.ListCount - 1
open list2.list(i) For Binary Access Write As #2 'list2.list(0)=c:\1.txt
a(i) = Space(LOF(2))
Get #2, , a(i)
close #2
Next

追问

Get #2, , a(i)
显示下标越界。

Answer 3

Dim arr() As Byte, a()
ReDim a(list2.ListCount - 1)
For i = 0 To list2.ListCount - 1
Open list2.List(i) For Binary Access Write As #2 'list2.list(0)=c:\1.txt
ReDim arr(LOF(2))
Get #2, , arr
Close #2
a(i) = arr
Next