VBScript 字符串替换为范围而不是字符串?

本文介绍了VBScript 字符串替换为范围而不是字符串?的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

Replace() 已经存在，但该函数将字符串作为参数.我需要范围.

Replace() already exists, but that function takes strings as parameters. I need range.

在我的字符串中有两个长度为 10 个字符的字符串".Greger 有 6 个字符和 4 个空格，另一个字符串有 10 个字符.

In my string there are two "strings" that are 10 characters long. Greger with 6 chars and 4 spaces and the other string with 10 characters.

"Greger AASSDDFFGG"

我想将 "Greger " 替换为 "googlioa "

我正在寻找的基本上是这样的:

What i'm looking for is basically this:

Replace(MyString,1,10) = "googlioa "

有什么办法可以实现吗?

Is there any way to achieve this?

推荐答案

如果要严格按位置替换，请使用 Left()、new 和 Mid() 的串联.让您开始:

If you want to replace strictly by position, use concatenation of Left(), new, and Mid(). To get you started:

>> Function replByPos(s, f, l, n) >> replByPos = Left(s, f-1) & n & Mid(s, f + l - 1) >> End Function >> s = "Greger AASSDDFFGG" >> r = replByPos(s, 1, 10, "googlioa ") >> WScript.Echo s >> WScript.Echo r >> Greger AASSDDFFGG googlioa AASSDDFFGG >>

进一步增强:

安全:f(rom) - 1 有风险 - 应该检查

填充新字符串 wrt l(ength)

也许您想在连接之前搜索 (Instr()) 以查找旧的 ("Greger ")

safety: f(rom) - 1 is risky - should be checked

padding of new string wrt l(ength)

perhaps you want to search (Instr()) for old ("Greger ") before the concatenation

再三考虑(偷了邦德的填充物):

也许我应该将 10 解释为 to/till/upto 值而不是长度/宽度规范.所以看看是否

Maybe I should have interpeted the 10 as a to/till/upto value instead of a length/width specification. So see whether

Option Explicit Function replByPos(src, from, till, ns) Dim w : w = till - from replByPos = Left(src, from - 1) & Left(ns & Space(w), w) & Mid(src, till) End Function Dim s : s = "Greger AASSDDFFGG" Dim ns : ns = "googlioa" WScript.Echo s WScript.Echo replByPos(s, 1, 10, ns) s = "Whatever Greger AASSDDFFGG" ns = "googlioa" Dim p : p = Instr(s, "Greger") WScript.Echo s WScript.Echo replByPos(s, p, p + 10, ns)

输出:

cscript 22811896.vbs Greger AASSDDFFGG googlioa AASSDDFFGG Whatever Greger AASSDDFFGG Whatever googlioa AASSDDFFGG

更符合您的规格.