如何用cmd.exe中的百分比字符替换？ [英] How to replace with percent character in cmd.exe?

问题描述

在Windows命令提示符下，我尝试用 set string = Hello World 替换字符串％20 。尝试使用字符串文字％20，如下所示：

In the Windows command prompt, I am trying to replace the space in set string=Hello World with the string %20. Naively trying to use the string literal %20 like this:

set string=%string: =%20%

结果 HelloWorld20％。尝试使用转义字符 ^ 这样：

results in HelloWorld20%. Trying to use the escape character ^ like this:

set string=%string: =^%20%

还会导致 HelloWorld20％。试图逃避％，将其加倍：

also results in HelloWorld20%. Trying to escape the % by doubling it like this:

set string=%string: =%%20%

结果 HelloWorld％20％ 。我也尝试使用另一个变量来做这样的替换：

results in HelloWorld%20%. I also tried to use another variable to do the replacement like this:

set r=%20
set string=%string: =%r%%

导致 HelloWorldr %% 。

我发现这个，它处理变量中百分号字符的转义。我还发现此，它处理转义的百分比字符输入。但是没有一个似乎适用于字符串替换。

I found this, which handles the escaping of percent characters in variables. I also found this, which handles the escaped input of percent characters. But neither one seems to apply to string replacing.

A

A tutorial/docu page for the Windows cmd.exe which I found online tells me I have the correct syntax, but does not cover replacing with percent characters.

阅读全部后，我尝试：

setlocal EnableDelayedExpansion
set string=!string: =%20!

导致！string：=％20！。

我没有想法，可以帮忙吗？

I am out of ideas, can you help?

推荐答案

在这种情况下，您需要延迟扩展

You need delayed expansion in this case:

set "str1=Hello World!"
set "str2=%20"
for /f "delims=" %a in ('cmd /v:on /c @echo "%str1: =!str2!%"') do set "str3=%~a"
echo %str3%

什么是延迟扩展？

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