dim函數第三個參數設置截取字符的長度問題

2019-10-26 18:05:17

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dim函數的第三個參數，也就是截取字符的長度，我在設置這個的時候，出了些問題:response.write mid(up_address,a(i),a(i+1)-1) & “ ”像我上面這樣寫的時候，它就會報錯，提示無效的過程調用或參數,
response.write mid(up_address,a(i),a(i+1)+1) & “ ”但是當我把其中的a(i+1)-1改為a(i+1)+1時，就能執行了,a(i+1)對應的值是11，可為什么只能減不能加呢?全部代碼如下:
VB code:
復制代碼代碼如下:

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) & " "
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) & " "
if a(i+1)=0 then
exit for
end if
response.write a(i+1) & " "
next

end if如上代碼，我是想把字符串按空格分解出來，但是mid的第三個參數那出了點問題，我本來是想這樣截取的：
VB code：
復制代碼代碼如下:

mid(up_address,a(i),a(i+1)-a(i)-1)
‘a(i)是空格的位置
‘a(i+1)是下一個空格的位置
‘a(i+1)-a(i)-1是兩個空格直間的字符長度

現在的問題是，經測試，mid的第三個參數那，無法使用減法，也就是說，我可以寫a(i+1)+，但不能寫a(i+1)-，想了好久，我一直不明白問題出在哪？應該怎么來解決呢？
出現這個問題是因為上面的MID函數的第三個參數出現了負數,下面是在網上找的測試的VBS代碼，原理一樣，如下的代碼：
VBScript code:
復制代碼代碼如下:

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) & " "
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)‘這里的結果為34，-1，所以導致出錯
MsgBox mid(up_address,a(i),a(i+1)-1) & " "
if a(i+1)=0 then
exit for
end if
MsgBox a(i+1) & " "
next
end if

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