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dim函数的第三个参数，也就是截取字符的长度，我在设置这个的时候，出了些问题,提示无效的过程调用或参数,本文详细介绍如何解决

dim函数的第三个参数，也就是截取字符的长度，我在设置这个的时候，出了些问题:response.write mid(up_address,a(i),a(i+1)-1) & “<br />”像我上面这样写的时候，它就会报错，提示无效的过程调用或参数,

response.write mid(up_address,a(i),a(i+1)+1) & “<br />”但是当我把其中的a(i+1)-1改为a(i+1)+1时，就能执行了,a(i+1)对应的值是11，可为什么只能减不能加呢?全部代码如下:

VB code:

代码如下:

dim a(),up_address

up_address = "aaa djaldk adflj adafadfasdfa afd ad"

redim a(len(up_address))

a(0) = instr(up_address," ")

response.write a(0) & "<br />"

if a(0)<>0 then

for i=0 to len(up_address)-1

a(i+1) = instr(a(i)+1,up_address," ")

response.write mid(up_address,a(i),a(i+1)-1) & "<br />"

if a(i+1)=0 then

exit for

end if

response.write a(i+1) & "<br />"

next

end if如上代码，我是想把字符串按空格分解出来，但是mid的第三个参数那出了点问题，我本来是想这样截取的：

VB code：

代码如下:

mid(up_address,a(i),a(i+1)-a(i)-1)

‘a(i)是空格的位置

‘a(i+1)是下一个空格的位置

‘a(i+1)-a(i)-1是两个空格直间的字符长度

现在的问题是，经测试，mid的第三个参数那，无法使用减法，也就是说，我可以写a(i+1)+，但不能写a(i+1)-，想了好久，我一直不明白问题出在哪？应该怎么来解决呢？

出现这个问题是因为上面的MID函数的第三个参数出现了负数,下面是在网上找的测试的VBS代码，原理一样，如下的代码：

VBScript code:

代码如下:

dim a(),up_address

up_address = "aaa djaldk adflj adafadfasdfa afd ad"

MsgBox len(up_address) '36

redim a(len(up_address)) 'a(36)

a(0) = instr(up_address," ")

MsgBox a(0) 'a(0)=4

MsgBox a(0) & "<br />"

if a(0)<>0 then

for i=0 to len(up_address)-1

a(i+1) = instr(a(i)+1,up_address," ")

MsgBox a(i) &" "& (a(i+1)-1)‘这里的结果为34，-1，所以导致出错

MsgBox mid(up_address,a(i),a(i+1)-1) & "<br />"

if a(i+1)=0 then

exit for

end if

MsgBox a(i+1) & "<br />"

next

end if