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:
实现代码如下：
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：
实现代码如下：
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:
实现代码如下：
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
以上就是【dim函数第三个参数设置截取字符的长度问题】的全部内容了,欢迎留言评论进行交流!